Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Circles Important Questions Very Short Answer Type to help strengthen their preparations for exams.

## TS Inter Second Year Maths 2B Circles Important Questions Very Short Answer Type

Question 1.

Find the centre and radius of the circle x^{2} + y^{2} + 6x + 8y – 96 = 0. (Mar. ’94)

Solution:

Given the equation of the circle is

x^{2} + y^{2} + 6x + 8y – 96 = 0 …….(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get 2g = 6 ⇒ g = 3

2f = 8 ⇒ f = 4

c = -96

Centre c = (-g, -f) = (-3, -4)

radius r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)

= \(\sqrt{(3)^2+(4)^2+96}\)

= √121

= 11

Question 2.

Find the centre and radius of the circle x^{2} + y^{2} + 2x – 4y – 4 = 0

Solution:

Given the equation of the circle is

x^{2} + y^{2} + 2x – 4y – 4 = 0 …….(1)

Comparing (1) with

x^{2} + y^{2} + 2gx + 2fy + c = 0

We get 2g = 2 ⇒ g = 1

2f = -4 ⇒ f = -2

c = -4

Centre c = (-g, -f) = (-1, 2)

radius r = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{(1)^2+(-2)^2+4}\)

= √9

= 3

Question 3.

Find the centre and radius of the circle x^{2} + y^{2} + 2ax – 2by + b^{2} = 0

Solution:

Given the equation of the circle is

x^{2} + y^{2} + 2ax – 2by + b2 = 0 …….(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get 2g = 2a ⇒ g = a

2f = -2b ⇒ f = -b

c = b^{2}

Centre c = (-g, -f) = (-a, b)

radius r = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{(a)^2+(-b)^2-b^2}\)

= √a^{2}

= a

Question 4.

Find the centre and radius of the circle 3x^{2} + 3y^{2} – 6x + 4y – 4 = 0.

Solution:

Given equation of the circle is 3x^{2} + 3y^{2} – 6x + 4y – 4 = 0

⇒ x^{2} + y^{2} – 2x + \(\frac{4}{3}\)y – \(\frac{4}{3}\) = 0 ………(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0

Question 5.

Find the centre and radius of the circle \(\sqrt{1+\mathrm{m}^2}\) (x^{2} + y^{2}) – 2cx – 2mcy = 0. (May ’10)

Solution:

Question 6.

If the circle x^{2} + y^{2} + ax + by – 12 = 0 has the centre at (2, 3), then find a, b, and the radius of the circle. (May ’09, ’07, Mar. ’08)

Solution:

Given the equation of the circle is

x^{2} + y^{2} + ax + by – 12 = 0 ……..(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0

we get 2g = a ⇒ g = \(\frac{a}{2}\)

2f = b ⇒ f = \(\frac{b}{2}\)

c = -12

Question 7.

If x^{2} + y^{2} + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3). Find g, f, and its radius. [(AP) May ’19]

Solution:

Given the equation of the circle is

x^{2} + y^{2} + 2gx + 2fy – 12 = 0 ……..(1)

Centre of (1) is C = (-g, -f)

Given that centre C = (2, 3)

∴ (-g, -f) = (2, 3)

⇒ g = -2, f = -3

Now Radius r = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{(-2)^2+(-3)^2-(-12)}\)

= √25

= 5

Question 8.

If x^{2} + y^{2} + 2gx + 2fy = 0 represents a circle with centre (-4, -3) then find g, f, and the radius of the circle. [(AP) May ’17]

Solution:

Given the equation of the circle is

x^{2} + y^{2} + 2gx + 2fy = 0 …….(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0

we get g = g, f = f, c = 0

Centre of (1) is C = (-g, -f)

Given that centre, C = (-4, -3)

∴ (-g, -f) = (-4, -3)

we get g = 4, f = 3

Radius r = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{(4)^2+(3)^2-0}\)

= √25

= 5

Question 9.

Find the value of ‘a’, if 2x^{2} + ay^{2} – 3x + 2y – 1 = 0 represents a circle, and also find its radius. [(AP) Mar. ’15]

Solution:

Given equation is

2x^{2} + ay^{2} – 3x + 2y – 1 = 0 …….(1)

Comparing (1) with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

We get a = 2, h = 0, b = a, g = \(\frac{-3}{2}\), f = 1, c = -1

Now, equation (1) represents a circle

then a = b

⇒ 2 = a

∴ a = 2

The equation of the circle is

2x^{2} + 2y^{2} – 3x + 2y – 1 = 0

\(x^2+y^2-\frac{3}{2} x+y-\frac{1}{2}=0\)

Comparing the above equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

Question 10.

If the circle x^{2} + y^{2} – 4x + 6y + a = 0 has a radius of 4 then find ‘a’. [(AP) Mar. ’16]

Solution:

Given the equation of the circle is

x^{2} + y^{2} – 4x + 6y + a = 0 …….(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get 2g = -4 ⇒ g = -2

2f = 6 ⇒ f = 3

c = a

Given that Radius, r = 4

⇒ \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=4\)

⇒ \(\sqrt{(-2)^2+(3)^2-a}=4\)

⇒ \(\sqrt{4+9-a}\) = 4

⇒ \(\sqrt{13-\mathrm{a}}\) = 4

Squaring on both sides

13 – a = 16

⇒ 13 – 16 = a

⇒ a = -3

∴ a = -3

Question 11.

If x^{2} + y^{2} – 4x + 6y + c = 0 represents a circle with a radius of 6 then find the value of ‘c’.

Solution:

Given equation of the circle is x^{2} + y^{2} – 4x + 6y + c = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0,

we get g = -2, f = 3, c = c

Given that radius r = 6

⇒ \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=6\)

⇒ \(\sqrt{(-2)^2+(3)^2-c}=6\)

⇒ \(\sqrt{4+9-\mathrm{c}}\)

⇒ \(\sqrt{13-c}=6\)

Squaring on both sides

13 – c = 36

⇒ 13 – 36 = c

⇒ c = -23

Question 12.

Find the equation of the circle passing through (-2, 3) having the centre at (0, 0).

Solution:

Given centre C(h, k) = (0, 0)

Let the given point A (-2, 3)

Since A(-2, 3) is a point on the circle.

Radius r = CA = \(\sqrt{(-2-0)^2+(3-0)^2}\) = √13

∴ The equation of the required circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – 0)^{2} + (y – 0)^{2} = (√13)^{2}

⇒ x^{2} + y^{2} = 13

⇒ x^{2} + y^{2} – 13 = 0

Question 13.

Find the equation of the circle passing through the origin and having the centre at (-4, -3). (Mar. ’04; May ’02)

Solution:

Given centre, C(h, k) = (-4, -3)

Let, the given point A(0, 0).

Since, A(0, 0) is a point on the circle.

∴ Radius, r = CA = \(\sqrt{(-4-0)^2+(-3-0)^2}\)

= √25

= 5

∴ The equation of the required circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x + 4)^{2}+ (y + 3)^{2} = (5)^{2}

⇒ x^{2} + 16 + 8x + y^{2} + 9 + 6y = 25

⇒ x^{2} + y^{2} + 8x + 6y + 25 = 25

⇒ x^{2} + y^{2} + 8x + 6y = 0

Question 14.

Find the equation of the circle passing through (3, 4) and having the centre at (-3, 4).

Solution:

Given centre C(h, k) = (-3, 4)

Let the given point A = (3, 4)

Since ‘A’ is the point on the circle.

∴ Radius, r = CA = \(\sqrt{(-3-3)^2+(4-4)^2}\)

= √36

= 6

∴ The equation of the required circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x + 3)^{2} + (y – 4)^{2} = 6^{2}

⇒ x^{2} + 9 + 6x + y^{2} + 16 – 8y = 36

⇒ x^{2} + y^{2} + 6x – 8y – 11 = 0

Question 15.

Find the equation of the circle which is concentric with x^{2} + y^{2} – 6x – 4y – 12 = 0 and passing through (-2, 14). [(TS) May ’17; Mar & May ’14]

Solution:

Given equation of the circle is x^{2} + y^{2} – 6x – 4y – 12 = 0

Equation of the circle concentric with the circle

x^{2} + y^{2} – 6x – 4y – 12 = 0 is x^{2} + y^{2} – 6x – 4y + k = 0 …….(1)

Equation (1) passes through (-2, 14) then

(-2)^{2} + (14)^{2} – 6(-2) – 4(14) + k = 0

⇒ 4 + 196 + 12 – 56 + k = 0

⇒ 156 + k = 0

⇒ k = -156

The required circle from (1) is x^{2} + y^{2} – 6x – 4y – 156 = 0.

Question 16.

Find the equation of the circle with (-4, 3), (3, -4) as ends of a diameter.

Solution:

Let A = (x_{1}, y_{1}) = (-4, 3), B = (x_{2}, y_{2}) = (3, -4) are the given points.

∴ The required equation of the circle is (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x + 4) (x – 3) + (y – 3) (y + 4) = 0

⇒ x^{2} – 3x + 4x – 12 + y^{2} + 4y – 3y – 12 = 0

⇒ x^{2} + y^{2} + x + y – 24 = 0

Question 17.

Find the other end of the diameter of the circle x^{2} + y^{2} – 8x – 8y + 27 = 0 if one end of it is (2, 3). [(AP) Mar. ’20; May ’12]

Solution:

Given the equation of the circle is x^{2} + y^{2} – 8x – 8y + 27 = 0.

Let A(2, 3) be the given point.

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

g = -4, f = -4, c = 27

Centre of the circle, C(h, k) = (-g, -f) = (4, 4)

Let, B(x, y) be the other end of the diameter

∵ ‘C’ is the midpoint of \(\overline{\mathrm{AB}}\) then (4, 4) = \(\left(\frac{2+x}{2}, \frac{3+y}{2}\right)\)

\(\frac{2+x}{2}\) = 4

⇒ 2 + x = 8

⇒ x = 8 – 2

⇒ x = 6

\(\frac{3+y}{2}\) = 4

⇒ 3 + y = 8

⇒ y = 8 – 3

⇒ y = 5

∴ The other end of the diameter is B = (6, 5).

Question 18.

Obtain the parametric equation of the circle represented by x^{2} + y^{2} = 4. (Mar. ’14)

Solution:

Given the equation of the circle is x^{2} + y^{2} = 4

This is of the form x^{2} + y^{2} = r^{2}

Centre C = (h, k) = (0, 0); Radius r = 2

∴ The parametric equations of the circle are

x = h + r cos θ = 0 + 2 cos θ = 2 cos θ

y = k + r sin θ = 0 + 2 sin θ = 2 sin θ

0 ≤ θ ≤ 2π

Question 19.

Find the parametric equations of the circle (x – 3)^{2} + (y – 4)^{2} = 8^{2}. [Mar. ’18, ’16 (AP)]

Solution:

Given equation of the circle is (x – 3)^{2} + (y – 4)^{2} = 8^{2}

This is of the form (x – h)^{2} + (y – k)^{2} = r^{2}

we get h = 3, k = 4, r = 8

∴ Centre (h, k) = (3, 4)

Radius r = 8

∴ The parametric equations are

x = h + r cos θ = 3 + 8 cos θ

y = k + r sin θ = 4 + 8 sin θ

0 ≤ θ ≤ 2π

Question 20.

Obtain the parametric equations of the circle represented by x^{2} + y^{2} + 6x + 8y – 96 = 0.

Solution:

Given the equation of the circle is

x^{2} + y^{2} + 6x + 8y – 96 = 0 …….(1)

Comparing (1) with

x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = 6 ⇒ g = 3

2f = 8 ⇒ f = 4

c = -96

Centre, C(h, k) = (-g, -f) = (-3, -4)

Radius, r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)

= \(\sqrt{3^2+4^2+96}\)

= √121

= 11

The parametric equations of the circle are

x = h + r cos θ = -3 + 11 cos θ

y = k + r sin θ = -4 + 11 sin θ

0 ≤ θ ≤ 2π

Question 21.

Find the parametric equations of the circle x^{2} + y^{2} – 6x + 4y – 12 = 0. [(AP) May ’15; Mar. ’10, ’06]

Solution:

Given the equation of the circle is

x^{2} + y^{2} – 6x + 4y – 12 = 0 ……….(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = -6 ⇒ g = -3

2f = 4 ⇒ f = 2

f = -12

Centre, C(h, k) = (-g, -f) = (3, -2)

Radius, r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)

= \(\sqrt{(-3)^2+(2)^2+12}\)

= √25

= 5

The parametric equations of the circle are

x = h + r cos θ = 3 + 5 cos θ

y = k + r sin θ = -2 + 5 sin θ

0 ≤ θ ≤ 2π

Question 22.

Find the length of the tangent from (3, 3) to the circle x^{2} + y^{2} + 6x + 18y + 26 = 0. (Mar. ’05)

Solution:

Given equation of the circle is x^{2} + y^{2} + 6x + 18y + 26 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

g = 3, f = 9, c = 26

Given point P(x_{1}, y_{1}) = (3, 3)

Length of the tangent = \(\sqrt{S_{11}}\)

= \(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\)

= \(\sqrt{(3)^2+(3)^2+2(3)(3)+2(9)(3)+26}\)

= \(\sqrt{9+9+18+54+26}\)

= √116

Question 23.

If the length of the tangent from (2, 5) to the circle x^{2} + y^{2} – 5x + 4y + k = 0 is √37, then find k. [(TS) May, Mar. ’18; (AP) May ’17]

Solution:

Given equation of the circle is x^{2} + y^{2} – 5x + 4y + k = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

g = \(\frac{-5}{2}\), f = 2, c = k

Let the given point P(x_{1}, y_{1}) = (2, 5)

Given that the length of the tangent = √37

∴ \(\sqrt{S_{11}}\) = √37

Question 24.

If the length of the tangent from (5, 4) to the circle x^{2} + y^{2} + 2ky = 0 is 1, then find ‘k’. [(TS) Mar. ’20, ’15, May 15; (AP) May ’18, Mar. ’15; May ‘16, ‘15]

Solution:

The given equation of the circle is x^{2} + y^{2} + 2ky = 0.

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0 we get

g = 0, f = k, c = 0

Let, the given point P(x_{1}, y_{1}) = (5, 4)

Given the length of the tangent = 1

\(\sqrt{S_{11}}\) = 1

⇒ S_{11} = 1

⇒ \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}=1\)

⇒ (5)^{2} + (4)^{2} + 2(0)(5) + 2k(4) + 0 = 1

⇒ 25 + 16 + 8k = 1

⇒ 8k = -40

⇒ k = \(\frac{-40}{8}\) = -5

Question 25.

Show that the line lx + my + n = 0 is normal to the circle S = 0 if and only if lg + mf = n.

Solution:

The straight line lx + my + n = 0 is normal to the circle.

S = x^{2} + y^{2} + 2gx + 2fy + c = 0

If the centre, C(-g, -f) lies on lx + my +n = 0 then

l(-g) + m(-f) + n = 0

⇒ -lg – mf + n = 0

⇒ lg + mf = n

Question 26.

Find the equation of the polar of (2, 3) with respect to the circle x^{2} + y^{2} + 6x + 8y – 96 = 0. [May ’96]

Solution:

Given equation of the circle is x^{2} + y^{2} + 6x + 8y – 96 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

we get g = 3, f = 4, c = -96

Let the given point P(x_{1}, y_{1}) = (2, 3)

The equation of polar P(2, 3) w.r.t the given circle is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(2) + y(3) + 3(x + 2) + 4(y + 3) – 96 = 0

⇒ 2x + 3y + 3x + 6 + 4y + 12 – 96 = 0

⇒ 5x + 7y – 78 = 0

Question 27.

Find the polar of (1, -2) with respect to x^{2} + y^{2} – 10x – 10y + 25 = 0. [(TS) Mar. ’15]

Solution:

Given equation of the circle is x^{2} + y^{2} – 10x – 10y + 25 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -5, f = -5, c = 25

Let the given point P(x_{1}, y_{1}) = (1, -2)

The equation of polar P(1, -2) w.r.t the given circle is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(1) + y(-2) – 5(x + 1) – 5(y – 2) + 25 = 0

⇒ x – 2y – 5x – 5 – 5y + 10 + 25 = 0

⇒ -4x – 7y + 30 = 0

⇒ 4x + 7y – 30 = 0

Question 28.

Find the value of k if the points (1, 3) and (2, k) are conjugate with respect to the circle x^{2} + y^{2} = 35. [(AP) May ’19, Mar. ’17; (TS) Mar. ’19, ’16, May ’17]

Solution:

Given the equation of the circle is x^{2} + y^{2} = 35

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

we get g = 0, f = 0, c = -35

Let, the given points are A(x_{1}, y_{1}) = (1, 3), B(x_{2}, y_{2}) = (2, k)

Since the given points are conjugate then S_{12} = 0

⇒ x_{1}x_{2}+ y_{1}y_{2}+ g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = 0

⇒ 1(2) + 300 + 0(1 + 2) + 0(3 + k) – 35 = 0

⇒ 2 + 3k – 35 = 0

⇒ 3k = 33

⇒ k = 11

Question 29.

If(4, k) and (2, 3) are conjugate points with respect to the circle x^{2} + y^{2} = 17 then find k.

Solution:

Given equation of the circle is x^{2} + y^{2} – 17 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

g = 0, f = 0, c = -17

Let, the given points be

A(x_{1}, y_{1}) = (4, k), B(x_{2}, y_{2}) = (2, 3)

Since the given points are conjugate then

S_{12} = 0

⇒ x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = o

⇒ 4(2) + k(3) + 0(4 + 2) + 0(k + 3) – 17 = 0

⇒ 8 + 3k + 0 + 0 – 17 = 0

⇒ 3k – 9 = 0

⇒ 3k = 9

⇒ k = 3

Question 30.

Find the value of ‘k’ if the points (4, 2) and (k, -3) are conjugate points with respect to the circle x^{2} + y^{2} – 5x + 8y + 6 = 0. [(AP) Mar. ’19; (TS) ’17; May ’14]

Solution:

Given equation of the circle is x^{2} + y2 – 5x + 8y + 6 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

g = \(\frac{-5}{2}\), f = 4, c = 6

Let the given points are

A(x_{1}, y_{1}) = (4, 2) and B(x_{2}, y_{2}) = (k, -3)

Since the given points are conjugate then S_{12} = 0

⇒ x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = 0

⇒ 4(k) + 2(-3) + \(\frac{-5}{2}\) (4 + k) + 4(2 – 3) + 6 = 0

⇒ 4k – 6 – 10 – \(\frac{5k}{2}\) – 4 + 6 = 0

⇒ \(\frac{-5k}{2}\) + 4k – 14 = 0

⇒ \(\frac{-5 k+8 k-28}{2}\) = 0

⇒ 3k – 28 = 0

⇒ 3k = 28

⇒ k = \(\frac{28}{3}\)

Question 31.

Find the die equation of the circle with centre (2, -3) and radius 4.

Solution:

Given centre c(h, k) = (2, -3)

radius r = 4

The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – 2)^{2} + (y + 3)^{2} = (4)^{2}

⇒ x^{2} + 4 – 4x + y^{2} + 9 + 6y = 16

⇒ x^{2} + y^{2} – 4x + 6y – 3 = 0

Question 32.

Find the equation of the circle with centre (-a, -b) and radius \(\sqrt{\mathbf{a}^2-\mathbf{b}^2}\).

Solution:

Given centre c(h, k) = (-a, -b)

radius r = \(\sqrt{\mathbf{a}^2-\mathbf{b}^2}\)

The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x + a)^{2} + (y + b)^{2} = \(\left(\sqrt{a^2-b^2}\right)^2\)

⇒ x^{2} + a^{2} + 2ax + y^{2} + b^{2} + 2by = a^{2} – b^{2}

⇒ x^{2} + y^{2} + 2ax + 2by + 2b^{2} = 0

Question 33.

Find the equation of the circle with centre c(a, -b) and radius a + b.

Solution:

Given centre c(h, k) = (a, -b)

radius r = a + b

The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – a)^{2} + (y + b)^{2} = (a + b)^{2}

⇒ x^{2} + a^{2} – 2ax + y^{2} + b^{2} + 2by = a^{2} + b^{2} + 2ab

⇒ x^{2} + y^{2} – 2ax + 2by – 2ab = 0

Question 34.

Find the centre and radius of the circle 3x^{2} + 3y^{2} – 5x – 6y + 4 = 0.

Solution:

Given equation of the circle is 3x^{2} + 3y^{2} – 5x – 6y + 4 = 0

Question 35.

Find the equation of the circle whose extremities of a diameter are (1, 2) and (4, 5).

Solution:

Let A(x_{1}, y_{1}) = (1, 2) and B(x_{2}, y_{2}) = (4, 5) be the two given points.

∴ The equation of the required circle is (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x – 1)(x – 4) + (y – 2)(y – 5) = 0

⇒ x^{2} – 4x – x + 4 + y^{2} – 2y – 5y + 10 = 0

⇒ x^{2} + y^{2} – 5x – 7y + 14 = 0

Question 36.

Find the position of the point P(3, 4) w.r.t the circle x^{2} + y^{2} – 4x – 6y – 12 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 4x – 6y – 12 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -2, f = -3, c = -12

Given point P(x_{1}, y_{1}) = (3, 4)

Now, S11 = \(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)

= (3)^{2} + (4)^{2} + 2(-2)(3) + 2(-3)(4) – 12

= 9 + 16 – 12 – 24 – 12

= 25 – 48

= -23

Since S_{11} < 0, then the point, P(3, 4) is inside the given circle.

Question 37.

Find the position of the point (1, 5) with respect to the circle x^{2} + y^{2} – 2x – 4y + 3 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 2x – 4y + 3 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -1, f = -2, c = 3

Given point P(x_{1}, y_{1}) = (1, 5)

Now S11 = \(\mathrm{x}_1{ }^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)

= (1)^{2} + (5)^{2} + 2(-1)(1) + 2(-2)(5) + 3

= 1 + 25 – 2 – 20 + 3

= 29 – 22

= 7

Since S_{11} > 0, then the point, P(1, 5) is outside the given circle.

Question 38.

Find the power of (1, 2) with respect to the circle x^{2} + y^{2} + 6x + 8y – 96 = 0.

Solution:

The given equation of the circle is x^{2} + y^{2} + 6x + 8y – 96 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

we get g = 3, f = 4, c = -96

Point P(x_{1}, y_{1}) = (1, 2)

∴ The power of P(1, 2) w.r.t the given circle is S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)

= (1)^{2} + (2)^{2} + 2(3)(1) + 2(4)(2) – 96

= 1 + 4 + 6 + 16 – 96

= -69

Question 39.

Find the power of the die point (-1, 1) with respect to the circle x^{2} + y^{2} – 6x + 4y – 12 = 0. [(TS) Mar. ’16]

Solution:

The given equation of the circle is x^{2} + y^{2} – 6x + 4y – 12 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

we get g = -3, f = 2, c = -12

Point P(x_{1}, y_{1}) = (-1, 1)

∴ The power of P(-1, 1) w.r.t the given circle is

S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)

= (-1)^{2} + (1)^{2} + 2(-3)(-1) + 2(2)(1) – 12

= 1 + 1 + 6 + 4 – 12

= 12 – 12

= 0

Question 40.

Find the length of the tangent from (2, 5) to the circle x^{2} + y^{2} – 5x + 4y – 5 = 0.

Solution:

The given equation of the circle is x^{2} + y^{2} – 5x + 4y – 5 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0,

we get g = \(\frac{-5}{2}\), f = 2, c = -5

Point, P(x_{1}, y_{1}) = (2, 5)

∴ The length of the tangent = \(\sqrt{S_{11}}\)

Question 41.

Find the length of the tangent from (-2, 5) to the circle x^{2} + y^{2} – 25 = 0. [(TS) May ’16]

Solution:

The given equation of the circle is x^{2} + y^{2} – 25 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

we get g = 0, f = 0, c = -25

Point P(x_{1}, y_{1}) = (-2, 5)

∴ The length of the tangent = \(\sqrt{S_{11}}\)

Question 42.

Find the length of the tangent from (12, 17) to the circle x^{2} + y^{2} – 6x – 8y – 25 = 0.

Solution:

The given equation of the circle is x^{2} + y^{2} – 6x – 8y – 25 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0,

We get g = -3, f = -4, c = -25

Point P(x_{1}, y_{1}) = (12, 17)

∴ The length of the tangent = \(\sqrt{S_{11}}\)

Question 43.

Find the equation of the tangent at P(-1, 1) of the circle x^{2} + y^{2} – 6x + 4y – 12 = 0. [(TS) Mar. ’16]

Solution:

Given equation of the circle is x^{2} + y^{2} – 6x + 4y – 12 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

we get g = -3, f = 2, c = -12

The given point P(x_{1}, y_{1}) = (-1, 1)

∴ The equation of the tangent at P is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(-1) + y(1) + (-3) (x – 1) + 2(y + 1) + c = 0

⇒ -x + y – 3x + 3 + 2y + 2 – 12 = 0

⇒ -4x + 3y – 7 = 0

⇒ 4x – 3y + 7 = 0

Question 44.

Find the area of the triangle formed by the tangent at P(x1, y1) to the circle x^{2} + y^{2} = a^{2} with the coordinate axes. [(AP) Mar. ’20; Apr. ’00, ’98, ’93]

Solution:

Given the equation of the circle is x^{2} + y^{2} = a^{2}

The equation of the tangent at P(x_{1}, y_{1}) to the given circle is S_{1} = 0.

Question 45.

Find the chord of contact of (1, 1) w.r.t the circle x^{2} + y^{2} = 9. [(AP) Mar. ’20]

Solution:

Given the equation of the circle is x^{2} + y^{2} = 9

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0,

We get g = 0, f = 0, c = -9

Given point P(x_{1}, y_{1}) = (1, 1)

∴ The equation of chord of contact is S1 = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(1) + y(1) + 0(x + 1) + 0(y + 1) – 9 = 0

⇒ x + y – 9 = 0

Question 46.

Find the pole of ax + by + c = 0 with respect to the circle x^{2} + y^{2} = r^{2}. [(AP) May ’16]

Solution:

Given the equation of the circle is x^{2} + y^{2} = r^{2}

Let P(x_{1}, y_{1}) be the pole of ax + by + c = 0 …….(1)

The polar of P w.r. t the circle x^{2} + y^{2} = r^{2} is S_{1} = 0

xx_{1} + yy_{1} – r^{2} = 0 ……(2)

Now (1) and (2) represent the same line

Question 47.

Discuss the relative position of the pair of circles (x – 2)^{2} + (y + 1)^{2} = 9, (x + 1)^{2} + (y – 3)^{2} = 4.

Solution:

Given equations of the circles are

(x – 2)^{2} + (y + 1)^{2} = 9 ……..(1)

(x + 1)^{2} + (y – 3)^{2} = 4 ………(2)

For circle (1),

Centre c_{1} = (2, -1); radius r_{1} = 3

For circle (2),

Centre c_{2} = (-1, 3); radius r_{2} = 2

c_{1}c_{2} = \(\sqrt{(2+1)^2+(-1-3)^2}\)

= \(\sqrt{9+16}\)

= √25

= 5

Now r_{1} + r_{2} = 3 + 2 = 5

∴ c_{1}c_{2} = r_{1} + r_{2}

∴ The given circles touch each other externally.

Question 48.

Find the number of common tangents that exist for the pair of circles x^{2} + y^{2} = 4, x^{2} + y^{2} – 6x – 8y + 16 = 0. (May ’93)

Solution:

Given equations of the circles are

x^{2} + y^{2} = 4 ……(1)

x^{2} + y^{2} – 6x – 8y + 16 = 0 ………(2)

For circle (1),

centre, C_{1} = (0, 0); Radius, r_{1} = 2

For circle (2),

centre, C_{2} = (3, 4)

Radius, r_{2} = \(\sqrt{(-3)^2+(-4)^2-16}\) = √9 = 3

C_{1}C_{2} = \(\sqrt{(0-3)^2+(0-4)^2}\) = √25 = 5

Now, r_{1} + r_{2} = 3 + 2 = 5

∴ C_{1}C_{2} = r_{1} + r_{2}

∴ The given circles touch each other externally.

∴ No. of common tangents = 3

Question 49.

Find the equation of the circle with centre (-7, -3) and radius ‘4’. (May ’93)

Solution:

Given centre, C(h, k) = (-7, -3)

radius, r = 4

∴ The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x + 7)^{2} + (y + 3)^{2} = 4^{2}

⇒ x^{2} + 49 + 14x + y^{2} + 9 + 6y = 16

⇒ x^{2} + y^{2} + 14x + 6y + 42 = 0

Question 50.

Find the equation of the circle passing through (2, -1) having the centre at (2, 3).

Solution:

Given centre, C(h, k) = (2, 3)

Let, the given point A(2, -1)

Since, A(2, -1) is a point on the circle

∴ Radius, r = CA = \(\sqrt{(2-2)^2+(-1-3)^2}\)

= \(\sqrt{0+(-4)^2}\)

= √16

= 4

∴ The equation of the required circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – 2)^{2} + (y – 3)^{2} = 4^{2}

⇒ x^{2} + 4 – 4x + y^{2} + 9 – 6y = 16

⇒ x^{2} + y^{2} – 4x – 6y – 3 = 0

Question 51.

Find the values of a, b if ax^{2} + bxy + 3y^{2} – 5x + 2y – 3 = 0 represents a circle. Also, find the radius and centre of the circle.

Solution:

Given equation is ax^{2} + bxy + 3y^{2} – 5x + 2y – 3 = 0 …….(1)

Comparing the equation (1) with ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

We get a = a,

2h = b ⇒ h = \(\frac{b}{2}\)

b = 3

Now, equation (1) represents a circle, then

(i) a = b ⇒ a = 3

(ii) h = 0 ⇒ \(\frac{b}{2}\) = 0 ⇒ b = 0

The equation of the circle is 3x^{2} + 3y^{2} – 5x + 2y – 3 = 0

⇒ \(x^2+y^2-\frac{5}{3} x+\frac{2}{3} y-1=0\)

Comparing the above equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

Question 52.

Find the equation of the circle with (4, 2), (1, 5) as ends of a diameter. [(AP) Mar. ’19]

Solution:

Let A(x_{1}, y_{1}) = (4, 2) and B(x_{2}, y_{2}) = (1, 5) are the two given points.

∴ The equation of the required circle is (x – x_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0

⇒ (x – 4) (x – 1) + (y – 2) (y – 5) = 0

⇒ x^{2} – 4x – x + 4 + y^{2} – 5y – 2y + 10 = 0

⇒ x^{2} + y^{2} – 5x – 7y + 14 = 0

Question 53.

Show that A(3, -1) lies on the circle x^{2} + y^{2} – 2x + 4y = 0. Also, find the other end of the diameter through A. (May ’99, ’95, ’91)

Solution:

Given equation of the circle is x^{2} + y^{2} – 2x + 4y = 0 ……(1)

Given point A = (3, -1)

Now substituting the point A (3, -1) in equation (1)

(3)^{2} + (-1)^{2} – 2(3) + 4 (-1) = 0

9 + 1 – 6 – 4 = 0

10 – 10 = 0

0 = 0

∴ The point A(3, -1) lies on the circle x^{2} + y^{2} – 2x + 4y = 0.

Comparing the equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -1, f = 2, c = 0

Centre C = (-g, -f) = (1, -2)

Let B = (x, y) be the other end of the diameter.

Since ‘C’ is the midpoint of \(\overline{\mathrm{AB}}\) then

C = \(\left[\frac{\mathrm{x}_1+\mathrm{x}_2}{2}, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}\right]\)

(1, -2) = \(\left(\frac{3+\mathrm{x}}{2}, \frac{-1+\mathrm{y}}{2}\right)\)

\(\frac{3+x}{2}\) = 1

⇒ 3 + x = 2

⇒ x = 2 – 3

⇒ x = -1

\(\frac{-1+\mathrm{y}}{2}\) = -2

⇒ -1 + y = -4

⇒ y = -4 + 1

⇒ y = -3

∴ Other end of the diameter B(x, y) = (-1, -3)

Question 54.

Find the equation of the circle passing through (3, -4) and concentric with x^{2} + y^{2} + 4x – 2y + 1 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} + 4x – 2y + 1 = 0

The equation of the circle concentric with the given circle is

x^{2} + y^{2} + 4x – 2y + k = 0 ……..(1)

The circle (1) passes through (3, -4) then

(3)^{2} + (-4)^{2} + 4(3) – 2(-4) + k = 0

⇒ 9 + 16 + 12 + 8 + k = 0

⇒ 45 + k = 0

⇒ k = -45

∴ The equation of the required circle is x^{2} + y^{2} + 4x – 2y – 45 = 0

Question 55.

Find the lengths of the intercepts made by the circle x^{2} + y^{2} + 8x – 12y – 9 = 0 on coordinate axes.

Solution:

Given equation of the circle is x^{2} + y^{2} + 8x – 12y – 9 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0,

we get g = 4, f = -6, c = -9

Length of the intercept made on X-axis = \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\)

= \(2 \sqrt{(4)^2+9}\)

= 2√25

= 2(5)

= 10

Length of the intercept made on Y-axis = \(2 \sqrt{\mathrm{f}^2-\mathrm{c}}\)

= \(2 \sqrt{(-6)^2+9}\)

= 2√45

= 6√5

Question 56.

Find the power of the point (3, 4) with respect to x^{2} + y^{2} – 4x – 6y – 12 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 4x – 6y- 12 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

g = -2, f = -3, c = -12

Given point P(x_{1}, y_{1}) = (3, 4)

Power of (3, 4) is S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\)

= (3)^{2} + (4)^{2} + 2(-2) (3) + 2(-3) (4) – 12

= 9 + 16 – 12 – 24 – 12

= -23

Question 57.

Find the polar of (1, 2) with respect to x^{2} + y^{2} = 7. (Mar. ’95)

Solution:

Given equation of the circle is x^{2} + y^{2} – 7 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

g = 0, f = 0, c = -7

Let the given point P(x_{1}, y_{1}) = (1, 2)

The equation of polar of P(1, 2) w.r.t. the given circle is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(1) + y(2) + 0(x + 1) + 0(y + 2) – 7 = 0

⇒ x + 2y – 7 = 0

Question 58.

Find the pole of the line 3x + 4y – 12 = 0 with respect to x^{2} + y^{2} = 24.

Solution:

Comparing 3x + 4y – 12 = 0 with the equation lx + my + n = 0, we get

l = 3, m = 4, n = -12

Comparing x^{2} + y^{2} = 24 with x^{2} + y^{2} = a^{2}

we get a^{2} = 24

∴ Pole = \(\left(\frac{-l \mathrm{a}^2}{\mathrm{n}}, \frac{-m \mathrm{a}^2}{\mathrm{n}}\right)\)

= \(\left(\frac{-3(24)}{-12}, \frac{-4(24)}{-12}\right)\)

= (6, 8)

Question 59.

Show that the points (-6, 1) and (2, 3) are conjugate points with respect to the circle x^{2} + y^{2} – 2x + 2y + 1 = 0. (Mar. ’96)

Solution:

Given equation of the circle is x^{2} + y^{2} – 2x + 2y + 1 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0,

we get g = -1, f = 1, c = 1

Let the given points are P(x_{1}, y_{1}) = (-6, 1), Q(x_{2}, y_{2}) = (2, 3)

Now S_{12} = x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c

= (-6)(2) + 1(3) – 1(-6 + 2) + 1(1 + 3) + 1

= 12 + 3 – (-4) + 1(4) + 1

= -12 + 3 + 4 + 4 + 1

= -12 + 12

= 0

Since S_{12} = 0 then, the given points are conjugate w.r.t. the given circle.

Question 60.

Find the pair of tangents drawn from (0, 0) to x^{2} + y^{2} + 10x + 10y + 40 = 0. (Mar. ’94)

Solution:

Given equation of the circle is x^{2} + y^{2} + 10x + 10y + 40 = 0

Here g = 5, f = 5, c = 40

Let the given point P(x_{1}, y_{1}) = (0, 0)

The equation of the pair of tangents drawn from (0, 0) to x^{2} + y^{2} + 10x + 10y + 40 = 0 is \(\mathrm{S}_1^2\) = SS_{11}

⇒ (xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c)^{2} = (x^{2} + y^{2} + 2gx + 2fy + c) \(\left(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\right)\)

⇒ [x(0) + y(0) + 5(x + 0) + 5(y + 0) + 40]^{2} = (x^{2} + y^{2} + 10x + 10y + 40) [0 + 0 + 2(5)(0) + 2(5)(0) + 40]

⇒ (5x + 5y + 40)^{2} = (x^{2} + y^{2} + 10x + 10y + 40)(40)

⇒ 25(x + y + 8)^{2} = (x^{2} + y^{2} + 10x + 10y + 40)(40)

⇒ 5(x^{2} + y^{2} + 64 + 2xy + 16y + 16x) = (x^{2} + y^{2} + 10x + 10y + 40) (8)

⇒ 5x^{2} + 5y^{2} + 320 + 10xy + 80y + 80x = 8x^{2} + 8y^{2} + 80x + 80y + 320

⇒ 3x^{2} – 10xy + 3y^{2} = 0

Question 61.

Find the equation to the pair of tangents drawn from (3, 2) to the circle x^{2} + y^{2} – 6x + 4y – 2 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 6x + 4y – 2 = 0

Here, g = -3, f = 2, c = -2

Let, the given point P(x_{1}, y_{1}) = (3, 2)

∴ The equation of the pair of tangents drawn from (3, 2) to the circle x^{2} + y^{2} – 6x + 4y – 2 = 0 is SS_{11} = \(\mathrm{S}_1^2\)

⇒ (x^{2} + y^{2} + 2gx + 2fy + c) \(\left(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{g}_1+2 \mathrm{fy} \mathrm{y}_1+\mathrm{c}\right)\) = (xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c)^{2}

⇒ (x^{2} + y^{2} – 6x + 4y – 2) [(3)^{2} + (2)^{2} + 2(-3)(3) + 2(2) (2) – 2] = [x(3) + y(2) – 3(x + 3) + 2(y + 2) + (-2)]^{2}

⇒ (x^{2} + y^{2} – 6x + 4y – 2) (9 + 4- 18 + 8- 2) = (3x + 2y – 3x – 9 + 2y + 4 – 2)^{2}

⇒ (x^{2} + y^{2} – 6x + 4y – 2) = (4y – 7)^{2}

⇒ x^{2} + y^{2} – 6x + 4y – 2 = 16y^{2} + 49 – 56y

⇒ x^{2} – 15y^{2} – 6x + 60y – 51 = 0

Question 62.

Find the pair of tangents from (4, 10) to the circle x^{2} + y^{2} = 25.

Solution:

Given the equation of the circle is x^{2} + y^{2} = 25

Here, g = 0, f = 0, c = -25

Let the given point P(x_{1}, y_{1}) = (4, 10)

The equation of the pair of tangents drawn from (4, 10) to x^{2} + y^{2} = 25 is SS_{11} = \(\mathrm{S}_1{ }^2\)

⇒ (x^{2} + y^{2} + 2gx + 2fy + c) \(\left(x_1^2+y_1^2+2 g x_1+2 f y_1+c\right)\) = [xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c]^{2}

⇒ (x^{2} + y^{2} – 25)((4)^{2} + (10)^{2} + 2(0)(4) + 2(0)(10) – 25) = [x(4) + y(10) + 0(x + 4) + 0(y + 10) – 25]^{2}

⇒ (x^{2} + y^{2} – 25) (16 + 100 – 25) = (4x + 10y – 25)^{2}

⇒ (x^{2} + y^{2} – 25) (91) = 16x^{2} + 100y^{2} + 625 + 80xy – 500y – 200x

⇒ 91x^{2} + 91y^{2} – 2275 = 16x^{2} + 100y^{2} + 625 + 80xy – 500y – 200x

⇒ 75x^{2} – 80xy – 9y^{2} + 200x + 500y – 2900 = 0

Question 63.

Find the angle between the pair of tangents drawn from (1, 3) to the circle x^{2} + y^{2} – 2x + 4y – 11 = 0. [(TS) Mar. ’16]

Solution:

Given equation of the circle is x^{2} + y^{2} – 2x + 4y – 11 = 0

Question 64.

m Find the equation of a circle with centre (1, 4) and radius 5. [Mar. ’17 (AP)]

Solution:

Given centre C(h, k) = (1, 4)

radius r = 5

∴ The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – 1)^{2} + (y – 4)^{2} = (5)^{2}

⇒ x^{2} + 1 – 2x + y^{2} + 16 – 8y = 25

⇒ x^{2} + y^{2} – 2x – 8y – 8 = 0

Question 65.

Obtain the parametric equation of the circle 4(x^{2} + y^{2}) = 9. [Mar. ’17 (TS)]

Solution:

Given equation of the circle is 4(x^{2} + y^{2}) = 9

⇒ x^{2} + y^{2} = \(\frac{9}{4}\)

This is of the form x^{2} + y^{2} = r^{2}

centre C = (h, k) = (0, 0)

radius r = \(\frac{3}{2}\)

∴ The parametric equations of the circle are

x = h + r cos θ = 0 + \(\frac{3}{2}\) cos θ = \(\frac{3}{2}\) cos θ

y = k + r sin θ = 0 + \(\frac{3}{2}\) sin θ = \(\frac{3}{2}\) sin θ

0 ≤ θ ≤ 2π

Question 66.

Find the centre and radius of the circle 3x^{2} + 3y^{2} + 6x – 12y – 1 = 0.

Solution:

Given equation of the circle is 3x^{2} + 3y^{2} + 6x – 12y – 1 = 0

⇒ x^{2} + y^{2} + 2x – 4y – \(\frac{1}{3}\) = 0 …….(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

Question 67.

Find the equation of the circle whose extremities of a diameter are (1, 2), (4, 6).

Solution:

Let A(x_{1}, y_{1}) = (1, 2) and B(x_{2}, y_{2}) = (4, 6) are the two given points.

∴ The equation of the required circle is (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x – 1) (x – 4) + (y – 2) (y – 6) = 0

⇒ x^{2} – 4x – x + 4 + y^{2} – 6y – 2y + 12 = 0

⇒ x^{2} + y^{2} – 5x – 8y + 16 = 0

Question 68.

Find the number of common tangents to the circles x^{2} + y^{2} + 6x + 6y + 14 = 0, x^{2} + y^{2} – 2x – 4y – 4 = 0.

Solution:

Given equations of the circles are

x^{2} + y^{2} + 6x + 6y + 14 = 0 ……..(1)

x^{2} + y^{2} – 2x – 4y – 4 = 0 ………(2)

For the circle (1), centre, C_{1} = (-3, -3)

Now, r_{1} + r_{2} = 2 + 3 = 5 = √25

∴ C_{1}C_{2} > r_{1} + r_{2}

∴ Number of common tangents = 4

Question 69.

Find the equation of the circle whose centre is (-1, 2) and which passes through (5, 6). [Mar. ’18 (TS)]

Solution:

Given centre c(h, k) = (-1, 2) and P(5, 6) be a point on the circle

Radius r = CP = \(\sqrt{(5+1)^2+(6-2)^2}\)

= \(\sqrt{36+16}\)

= √52

Equation of the required circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x + 1)^{2} + (y – 2)^{2} = (√52)^{2}

⇒ x^{2} + 1 + 2x + y^{2} + 4 – 4y = 52

⇒ x^{2} + y^{2} + 2x – 4y – 47 = 0

Question 70.

Write the parametric equations of the circle 2x^{2} + 2y^{2} = 7. [Mar. ’19 (TS)]

Solution:

Given the equation of the circle is 2x^{2} + 2y^{2} = 7

⇒ x^{2} + y^{2} = \(\frac{7}{2}\)

This is of the form x^{2} + y^{2} = r^{2}

centre C = (h, k) = (0, 0)

radius r = \(\frac{\sqrt{7}}{\sqrt{2}}\)

∴ The parametric equations of the circle are

x = h + r cos θ = 0 + \(\frac{\sqrt{7}}{\sqrt{2}}\) cos θ = \(\frac{\sqrt{7}}{\sqrt{2}}\) cos θ

y = k + r sin θ = 0 + \(\frac{\sqrt{7}}{\sqrt{2}}\) sin θ = \(\frac{\sqrt{7}}{\sqrt{2}}\) sin θ

0 ≤ θ ≤ 2π

Question 71.

Prove that the equation of the circle with centre C(h, k) and radius r is (x – h)^{2} + (y – k)^{2} = r^{2}.

Solution:

Let P(x, y) be a point P lies in the circle then PC = r

⇒ PC = r

⇒ \(\sqrt{(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2}\) = r

squaring on both sides

⇒ \(\left(\sqrt{(x-h)^2+(y-k)^2}\right)^2=r^2\)

⇒ (x – h)^{2} + (y – k)^{2} = r^{2}

∴ The locus of P is (x – h)^{2} + (y – k)^{2} = r^{2}

∴ The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

Question 72.

Prove that the equation of the circle with centre O(0, 0) and radius ‘r’ is x^{2} + y^{2} = r^{2}.

Solution:

Let P(x, y) be a point.

P lies in the circle then PO = r

\(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\) = r

Squaring on both sides

∴ x^{2} + y^{2} = r^{2}

Question 73.

Find the equation of the circle with centre (cos α, sin α) and radius 1.

Solution:

Given centre C(h, k) = (cos α, sin α)

radius r = 1

∴ The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – cos α)^{2} + (y – sin α)^{2} = (1)^{2}

⇒ x^{2} + cos^{2}α – 2x cos α + y^{2} + sin^{2}α – 2y sin α = 1

⇒ x^{2} + y^{2} – 2x cos α – 2y sin α + 1 – 1 = 0

⇒ x^{2} + y^{2} – 2x cos α – 2y sin α = 0

Question 74.

Find the equation of the circle with centre \(\left(\frac{-1}{2},-9\right)\) and radius 5.

Solution:

Given centre C(h, k) = \(\left(\frac{-1}{2},-9\right)\)

radius r = 5

∴ The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ \(\left(\mathrm{x}+\frac{1}{2}\right)^2+(\mathrm{y}+9)^2=(5)^2\)

⇒ x^{2} + x + \(\frac{1}{4}\) + y^{2} + 81 + 18y – 25 = 0

⇒ x^{2} + y^{2} + x + 18y + 56 + \(\frac{1}{4}\) = 0

⇒ 4x^{2} + 4y^{2} + 4x + 72y + 1 + 224 = 0

⇒ 4x^{2} + 4y^{2} + 4x + 72y + 225 = 0

Question 75.

Find the equation of the circle with centre \(\left(\frac{5}{2}, \frac{-4}{3}\right)\) and radius 6.

Solution:

Given centre C(h, k) = \(\left(\frac{5}{2}, \frac{-4}{3}\right)\)

radius r = 6

∴ The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ \(\left(x-\frac{5}{2}\right)^2+\left(y+\frac{4}{3}\right)^2=(6)^2\)

⇒ \(x^2+\frac{25}{4}-5 x+y^2+\frac{16}{9}+\frac{8 y}{3}=36\)

⇒ \(\frac{36 x^2+225-180 x+36 y^2+64+96 y}{36}\) = 36

⇒ 36x^{2} + 36y^{2} – 180x + 96y + 289 = 1296

⇒ 36x^{2} + 36y^{2} – 180x + 96y – 1007 = 0

Question 76.

Find the equation of the circle with centre (1, 7) and radius \(\frac{5}{2}\).

Solution:

Given centre C(h, k) = (1, 7)

radius r = \(\frac{5}{2}\)

∴ The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – 1)^{2} + (y – 7)^{2} = \(\left(\frac{5}{2}\right)^2\)

⇒ x^{2} + 1 – 2x + y^{2} + 49 – 14y = \(\frac{25}{4}\)

⇒ 4x^{2} + 4 – 8x + 4y^{2} + 196 – 56y = 25

⇒ 4x^{2} + 4y^{2} – 8x – 56y + 175 = 0

Question 77.

Find the equation of the circle with centre (0, 0) and radius 9.

Solution:

Given centre C(h, k) = (0, 0)

radius r = 9

∴ The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – 0)^{2} + (y – 0)^{2} = (9)^{2}

⇒ x^{2} + y^{2} – 81 = 0

Question 78.

Find the centre and radius of the circle 2x^{2} + 2y^{2} – 3x + 2y – 1 = 0.

Solution:

Given equation of the circle is 2x^{2} + 2y^{2} – 3x + 2y – 1 = 0

Question 79.

Find the equation of the circle having (7, -3), (3, 5) as the endpoints of a diameter.

Solution:

Let A(x_{1}, y_{1}) = (7, -3) and B(x_{2}, y_{2}) = (3, 5) are the two given points.

∴ The equation of the required circle is (x – x_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0

⇒ (x – 7)(x – 3) + (y + 3)(y – 5) = 0

⇒ x^{2} – 7x – 3x + 21 + y^{2} – 5y + 3y – 15 = 0

⇒ x^{2} + y^{2} – 10x – 2y + 6 = 0

Question 80.

Find the equation of the circle having (1, 1), (2, -1) as the endpoints of a diameter.

Solution:

Let A(x_{1}, y_{1}) = (1, 1) and B(x_{2}, y_{2}) = (2, -1) are the two given points.

∴ The equation of the required circle is (x – x_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0

⇒ (x – 1) (x – 2) + (y – 1) (y + 1) = 0

⇒ x^{2} – 2x – x + 2 + y^{2} – 1 = 0

⇒ x^{2} + y^{2} – 3x + 1 = 0

Question 81.

Find the equation of the circle having (0, 0), (8, 5) as the endpoints of a diameter.

Solution:

Let A(x_{1}, y_{1}) = (0, 0) and B(x_{2}, y_{2}) = (8, 5) are the two given points.

The equation of the required circle is (x – x_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0

⇒ (x – 0) (x – 8) + (y – 0) (y – 5) = 0

⇒ x^{2} – 8x + y^{2} – 5y = 0

⇒ x^{2} + y^{2} – 8x – 5y = 0

Question 82.

Obtain the parametric equations of the circle x^{2} + y^{2} – 4x – 6y – 12 = 0.

Solution:

Given the equation of the circle is

x^{2} + y^{2} – 4x – 6y – 12 = 0 ……..(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get 2g = -4 ⇒ g = -2

2f = -6 ⇒ f = -3

c = -12

Centre C(h, k) = (-g, -f) = (2, 3)

Radius r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)

= \(\sqrt{(-2)^2+(-3)^2+12}\)

= √25

= 5

∴ The parametric equations of the circle are

x = h + r cos θ = 2 + 5 cos θ

y = k + r sin θ = 3 + 5 sin θ, 0 ≤ θ ≤ 2π.

Question 83.

Show that the power of a point P(x1, y1) w.r.t the circle S = 0 is S11.

Solution:

Let S = x^{2} + y^{2} + 2gx + 2fy + c = 0 be the given circle and P(x_{1}, y_{1}) be any point in the plane, then centre C(-g, -f),

Question 84.

Prove that the length of the tangent drawn from an external point P(x_{1}, y_{1}) to the circle S = 0 is \(\sqrt{\mathbf{S}_{11}}\).

Solution:

Let S = x^{2} + y^{2} + 2gx + 2fy + c = 0 be the given circle

Centre C = (-g, -f) and Radius r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)

Let the tangents drawn from P touches the circle at A.

∴ ∠PAC = 90°

⇒ OP^{2} = PA^{2} + CA^{2}

Question 85.

Locate the position of the point P(1, 5) with respect to the circle x^{2} + y^{2} – 2x – 4y + 3 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 2x – 4y + 3 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0,

We get g = -1, f = -2, c = 3

Given point P(x_{1}, y_{1}) = (1, 5)

Now S_{11} = \(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)

= (1)^{2} + (5)^{2} + 2(-1)(1) + 2(-2)(5) + 3

= 1 + 25 – 2 – 20 + 3

= 29 – 22

= 7

Since S_{11} > 0, then the point P(1, 5) is outside the given circle.

Question 86.

Locate the position of the point P(4, 2) with respect to the circle 2x^{2} + 2y^{2} – 5x – 4y – 3 = 0.

Solution:

Given equation of the circle is 2x^{2} + 2y^{2} – 5x – 4y – 3 = 0

⇒ \(x^2+y^2-\frac{5}{2} x-2 y-\frac{3}{2}=0\)

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

= 20 – 14 – \(\frac{3}{2}\)

= 6 – \(\frac{3}{2}\)

= \(\frac{9}{2}\) > 0

Since S_{11} > 0, then the point P(4, 2) lies outside the circle.

Question 87.

Locate the position of the point P(2, -1) w.r.t the circle x^{2} + y^{2} – 2x – 4y + 3 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 2x – 4y + 3 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -1, f = -2, c = 3

Given point P(x_{1}, y_{1}) = (2, -1)

Now S_{11} = \(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\)

= (2)^{2} – (-1)^{2} + 2(-1)(2) + 2(-2)(-1) + 3

= 4 + 1 – 4 + 4 + 3

= 8 > 0

∵ S_{11} > 0, then the point P(2, -1) lies outside the circle.

Question 88.

Locate the position of the point (2, 4) w.r.t the circle x^{2} + y^{2} – 4x – 6y + 11 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 4x – 6y + 11 = 0

Comparing this equation of the circle is x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -2, f = -3, c = 11

Given point P(x_{1}, y_{1}) = (2, 4)

Now S_{11} = \(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)

= (2)^{2} + (4)^{2} + 2(-2)(2) + 2(4)(-3) + 11

= 4 + 16 – 8 – 24 + 11

= 31 – 32

= -1 < 0

∵ S_{11} < 0, then the point P(2, 4) lies inside the given circle.

Question 89.

Find the power of the point (5, -6) with respect to the circle x^{2} + y^{2} + 8x + 12y + 15 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} + 8x + 12y + 15 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = 4, f = 6, c = 15

Point P(x_{1}, y_{1}) = (5, -6)

∴ The power of P(5, -6) w.r.t the given circle is S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\)

S_{11} =(5)^{2} + (-6)^{2} + 2(4)(5) + 2(6)(-6) + 15

= 25 + 36 + 40 – 72 + 15

= 116 – 72

= 44

Question 90.

Find the power of the point (2, 3) with respect to the circle x^{2} + y^{2} – 2x + 8y – 23 = 0.

Solution:

The given equation of the circle is x^{2} + y^{2} – 2x + 8y – 23 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -1, f = 4, c = -23

Point P(x_{1}, y_{1}) = (2, 3)

∴ The power of P(2, 3) w.r.t the given circle is S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{fy}_1+\mathrm{c}\)

S_{11} = (2)^{2} – (3)^{2} + 2(-1)(2) + 2(4)(3) – 23

= 4 + 9 – 4 + 24 – 23

= 10

Question 91.

Find the power of the point (2, 4) with respect to the circle x^{2} + y^{2} – 4x – 6y – 12 = 0.

Solution:

The given equation of the circle is x^{2} + y^{2} – 4x – 6y – 12 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -2, f = -3, c = -12

Point P(x_{1}, y_{1}) = (2, 4)

∴ The power of P(2, 4) w.r.t the given circle is S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\)

S_{11} = (2)^{2} + (4)^{2} + 2(-2)(2) + 2(-3)(4) – 12

= 4 + 16 – 8 – 24 – 12

= -24

Question 92.

Find the length of tangent from P(-2, 5) to the circle x^{2} + y^{2} – 25 = 0.

Solution:

The given equation of the circle is x^{2} + y^{2} – 25 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = 0, f = 0, c = -25

Point P(x_{1}, y_{1}) = (-2, 5)

∴ The length of the tangent = \(\sqrt{\mathrm{S}_{11}}\)

Question 93.

Find the equation of the tangent at (7, -5) of the circle x^{2} + y^{2} – 6x + 4y – 12 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 6x + 4y – 12 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -3, f = 2, c = -12

The given point P(x_{1}, y_{1}) = (7, -5)

∴ The equation of the tangent at P is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(7) + y(-5) – 3(x + 7) + 2(y – 5) – 12 = 0

⇒ 7x – 5y – 3x – 21 + 2y – 10 – 12 = 0

⇒ 4x – 3y – 43 = 0

Question 94.

Find the equation of the tangent at (3, 4) of the circle x^{2} + y^{2} – 4x – 6y + 11 = 0.

Solution:

The given equation of the circle is x^{2} + y^{2} – 4x – 6y + 11 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -2, f = -3, c = 11

The given point P(x_{1}, y_{1}) = (3, 4)

The equation of the tangent at P is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(3) + y(4) + (-2)(x + 3) – 3(y + 4) + 11 = 0

⇒ 3x + 4y – 2x – 6 – 3y -12 + 11 = 0

⇒ x + y – 7 = 0

Question 95.

Find the equation of the normal at (3, -4) of the circle x^{2} + y^{2} + x + y – 24 = 0.

Solution:

The given equation of the circle is x^{2} + y^{2} + x + y – 24 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = \(\frac{1}{2}\), f = \(\frac{1}{2}\), c = -24

The given point P(x_{1}, y_{1}) = (3, -4)

∴ The equation of the tangent at P is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(3) + y(-4) + \(\frac{1}{2}\)(x + 3) + \(\frac{1}{2}\)(y – 4) – 24 = 0

⇒ 3x – 4y + \(\frac{x}{2}+\frac{3}{2}+\frac{y}{2}\) – 2 – 24 = 0

⇒ 6x – 8y + x + 3 + y – 4 – 48 = 0

⇒ 7x – 7y – 49 = 0

⇒ x – y – 7 = 0

The slope of the tangent is m = \(\frac{-a}{b}=\frac{-1}{-1}\) = 1

The slope of the normal at P = \(\frac{-1}{\mathrm{~m}}=\frac{-1}{1}\) = -1

∴ The equation of the normal at P is

\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)

⇒ y + 4 = -1(x – 3)

⇒ y + 4 = -x + 3

⇒ x + y + 1 = 0

Question 96.

Find the equation of the normal at (3, 5) of the circle x^{2} + y^{2} – 10x – 2y + 6 = 0. [(AP) Mar. ’18]

Solution:

The given equation of the circle is x^{2} + y^{2} – 10x – 2y + 6 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -5, f = -1, c = 6

The given point P(x_{1}, y_{1}) = (3, 5)

∴ The equation of the tangent at P is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(3) + y(5) + (-5)(x + 3) – 1(y + 5) + 6 = 0

⇒ 3x + 5y – 5x – 15 – y – 5 + 6 = 0

⇒ -2x + 4y – 14 = 0

⇒ x – 2y + 7 = 0

The slope of the tangent at P is

m = \(\frac{-a}{b}=\frac{-1}{-2}\) = 1

The slope of the normal at

P = \(\frac{-1}{m}=\frac{-1}{\frac{1}{2}}\) = -2

∴ The equation of the normal at P(3, 5) is

\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)

⇒ y – 5 = -2(x – 3)

⇒ y – 5 = -2x + 6

⇒ 2x + y – 11 = 0

Question 97.

Find the equation of the normal at (1, 3) of the circle 3(x^{2} + y^{2}) – 19x – 29y + 76 = 0.

Solution:

The given equation of the circle is 3x^{2} + 3y^{2} – 19x – 29y + 76 = 0

⇒ \(x^2+y^2-\frac{19}{3} x-\frac{29}{3} y+\frac{76}{3}=0\)

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get \(\mathrm{g}=\frac{-19}{6}, \quad \mathrm{f}=\frac{-29}{6}, \mathrm{c}=\frac{76}{3}\)

The given point P(x_{1}, y_{1}) = (1, 3)

∴ The equation of the tangent at P is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ \(x(1)+y(3)-\frac{19}{6}(x+1)-\frac{29}{6}(y+3)+\frac{76}{3}=0\)

⇒ 6x + 18y – 19x – 19 – 29y – 87 + 152 = 0

⇒ -13x – 11y + 46 = 0

⇒ 13x + 11y – 46 = 0

The slope of the tangent at P is

m = \(\frac{-a}{b}=\frac{-13}{11}\)

The slope of the normal at

P = \(\frac{-1}{\mathrm{~m}}=\frac{-1}{\frac{-13}{11}}=\frac{11}{13}\)

∴ The equation of the normal at P is

\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)

⇒ y – 3 = \(\frac{11}{33}\) (x – 1)

⇒ 13y – 39 = 11x – 11

⇒ 11x – 13y + 28 = 0

Question 98.

Find the equation of the normal at (1, 2) of the circle x^{2} + y^{2} – 22x – 4y + 25 = 0.

Solution:

The given equation of the circle is x^{2} + y^{2} – 22x – 4y + 25 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = -11, f = -2, c = 25

The given point P(x_{1}, y_{1}) = (1, 2)

∴ The equation of the tangent at P is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(1) + y(3) – 11(x + 1) – 2(y + 2) + 25 = 0

⇒ x + 2y – 11x – 11 – 2y – 4 + 25 = 0

⇒ -10x + 10 = 0

⇒ x – 1 = 0

The slope of the tangent at P is m = \(\frac{-1}{0}\)

The slope of the normal at

P = \(\frac{-1}{m}=\frac{-1}{\frac{-1}{0}}=0\)

∴ The equation of the normal at P(1, 2) is

\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)

⇒ y – 2 = 0(x – 1)

⇒ y – 2 = 0

Question 99.

Find the equation of the circle with centre (-3, 4) and touching y-axis.

Solution:

Given centre C(h, k) = (-3, 4).

The equation of the y-axis is x = 0.

Since the required circle is touching the y-axis, the y-axis is the tangent to the required circle.

∴ Radius r = The perpendicular distance from the centre C(-3, 4) to the y-axis (x = 0)

∴ r = d = \(\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}\)

= \(\frac{|1(-3)+0(4)+0|}{\sqrt{(1)^2+(0)^2}}=\frac{|-3|}{1}\)

= 3

∴ The required equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x + 3)^{2} + (y – 4)^{2} = (3)^{2}

⇒ x^{2} + 9 + 6x + y^{2} + 16 – 8y = 9

⇒ x^{2} + y^{2} + 6x – 8y + 16 = 0

Question 100.

Find the equation of tangent and normal at (3, 2) of the circle x^{2} + y^{2} – x – 3y – 4 = 0. [(AP) May ’19]

Solution:

The given equation of the circle is x^{2} + y^{2} – x – 3y – 4 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = \(\frac{-1}{2}\), f = \(\frac{-3}{2}\), c = -4

The given point P(x_{1}, y_{1}) = (3, 2)

∴ The equation of the tangent at P is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(3) + y(2) – \(\frac{1}{2}\)(x + 3) – \(\frac{3}{2}\)(y + 2) – 4 = 0

⇒ 6x + 4y – x – 3 – 3y – 6 – 8 = 0

⇒ 5x + y – 17 = 0

The slope of the tangent at P is

m = \(\frac{-5}{1}\) = -5

The slope of the normal at

P = \(\frac{-1}{m}=\frac{-1}{-5}=\frac{1}{5}\)

∴ The equation of the normal at P is

\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)

⇒ -2 = \(\frac{1}{5}\) (x – 3)

⇒ 5y – 10 = x – 3

⇒ x – 5y + 7 = 0

Question 101.

Find the equation of tangent and normal at (1, 1) of the circle 2x^{2} + 2y^{2} – 2x – 5y + 3 = 0.

Solution:

The given equation of the circle is 2x^{2} + 2y^{2} – 2x – 5y + 3 = 0

⇒ \(x^2+y^2-x-\frac{5}{2} y+\frac{3}{2}=0\) …..(1)

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get \(\mathrm{g}=\frac{-1}{2}, \mathrm{f}=\frac{-5}{4}, \mathrm{c}=-\frac{3}{2}\)

The given point P(x_{1}, y_{1}) = (1, 1)

∴ The equation of the tangent at P is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(1) + y(1) – \(\frac{1}{2}\)(x + 1) – \(\frac{5}{4}\)(y + 1) + \(\frac{3}{2}\) = 0

⇒ 4x + 4y – 2x – 2 – 5y – 5 + 6 = 0

⇒ 2x – y – 1 = 0

The slope of the tangent at P is m = \(\frac{-2}{-1}\) = 2

The slope of the normal at P is \(\frac{-1}{\mathrm{~m}}=\frac{-1}{2}\)

∴ The equation of the normal at P is

\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)

⇒ \(y-1=\frac{-1}{2}(x-1)\)

⇒ 2y – 2 = -x + 1

⇒ x + 2y – 3 = 0

Question 102.

Show that the circle S = x^{2} + y^{2} + 2gx + 2fy + c = 0 touches the

(i) X-axis if g^{2} = c

(ii) Y-axis if f^{2} = c

Solution:

Given the equation of the circle is

S = x^{2} + y^{2} + 2gx + 2fy + c = 0

(i) The intercept made by S = 0 on X-axis = \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\)

Since, the circle touches the X-axis, then

\(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\) = 0

⇒ \(\sqrt{g^2-c}\) = 0

⇒ g^{2} – c = 0

⇒ g^{2} = c

(ii) The intercept made by S = 0 on Y-axis is \(2 \sqrt{\mathrm{f}^2-\mathrm{c}}\)

Since, the circle touches the Y-axis, then

⇒ \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\)

⇒ \(\sqrt{f^2-c}\) = 0

⇒ f^{2} – c = 0

⇒ f^{2} = c

Question 103.

If the parametric values of two points A and B lying on the circle x^{2} + y^{2} – 6x + 4y – 12 = 0 are 30° and 60° respectively, then find the equation of the chord joining A and B.

Solution:

Given equation of the circle is x^{2} + y^{2} – 6x + 4y – 12 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

⇒ 2x – 6 + 2y + 4 = 5√3 + 5

⇒ 2x + 2y – 2 – 5 – 5√3 = 0

⇒ 2x + 2y – (7 + 5√3) = 0

Question 104.

Find the chord of contact of (0, 5) w.r.t the circle x^{2} + y^{2} – 5x + 4y – 2 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 5x + 4y – 2 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = \(\frac{-5}{2}\), f = 2, c = -2

Let, the given point P(x_{1}, y_{1}) = (0, 5)

∴ The equation of the chord of contact is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(0) + y(5) – \(\frac{5}{2}\)(x + 0) + 2(y + 5) – 2 = 0

⇒ 0 + 5y – \(\frac{5}{2}\)x + 2y + 10 – 2 = 0

⇒ 10y – 5x + 4y + 20 – 4 = 0

⇒ -5x + 14y + 16 = 0

⇒ 5x – 14y – 16 = 0

Question 105.

Find the chord of contact of (2, 5) w.r.t the circle x^{2} + y^{2} – 5x + 4y – 2 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 5x + 4y – 2 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = \(\frac{-5}{2}\), f = 2, c = -2

Let, the given point P(x_{1}, y_{1}) = (2, 5)

∴ The equation of the chord of contact is S_{1} = 0

⇒ xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(2) + y(5) – \(\frac{5}{2}\)(x + 2) + 2(y + 5) – 2 = 0

⇒ 2x + 5y – \(\frac{5}{2}\)x – 5 + 2y + 10 – 2 = 0

⇒ 2x – \(\frac{5}{2}\)x + 7y + 3 = 0

⇒ 4x – 5x + 14y + 6 = 0

⇒ -x + 14y + 6 = 0

⇒ x – 14y – 6 = 0

Question 106.

Show that the points (4, 2) and (3, -5) are conjugate points w.r.t the circle x^{2} + y^{2} – 3x – 5y + 1 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 3x – 5y + 1 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get \(\mathrm{g}=\frac{-3}{2}, \mathrm{f}=\frac{-5}{2}, \mathrm{c}=1\)

Let, the given point are A(x_{1}, y_{1}) = (4, 2), B(x_{2}, y_{2}) = (3, -5)

Now S_{12} = x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = 0

Since S_{12} = 0, the given points are conjugate w.r.t the given circle.

Question 107.

Find the value of ‘k’ if the points (1, 3) and (2, k) are conjugate w.r.t the circle x^{2} + y^{2} = 35.

Solution:

Given the equation of the circle is x^{2} + y^{2} = 35

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = 0, f = 0, c = -35

Let, the given points are A(x_{1}, y_{1}) = (1, 3), B(x_{2}, y_{2}) = (2, k)

Since the given points are conjugate, then S_{12} = 0

⇒ x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = 0

⇒ 1(2) + 3(k) + 0(1 + 2) + 0(3 + k) – 35 = 0

⇒ 2 + 3k + 0 + 0 – 35 = 0

⇒ 3k – 33 = 0

⇒ 3k = 33

⇒ k = 11

Question 108.

Find the value of ‘k’ if the points (4, 2) and (k, -3) are conjugate points w.r.t the circle x^{2} + y^{2} – 5x + 8y + 6 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 5x + 8y + 6 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

We get g = \(\frac{-5}{2}\), f = 4, c = 6

Let, the given points are A(x_{1}, y_{1}) = (4, 2) and B(x_{2}, y_{2}) = (k, -3)

Since the given points are conjugate, then S_{12} = 0

⇒ x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = 0

⇒ 4(k) + 2(-3) – \(\frac{5}{2}\)(4 + k) + 4(2 – 3) + 6 = 0

⇒ 4k – 6 – \(\frac{20}{2}-\frac{5 \mathrm{k}}{2}\) – 4 + 6 = 0

⇒ \(\frac{-5 k}{2}\) + 4k – 14 = 0

⇒ \(\frac{-5 k+8 k-28}{2}\) = 0

⇒ 3k – 28 = 0

⇒ 3k = 28

⇒ k = \(\frac{28}{3}\)

Question 109.

Find the angle between the tangents drawn from (3, 2) to the circle x^{2} + y^{2} – 6x + 4y – 2 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 6x + 4y – 2 = 0

Question 110.

Discuss the relative position of the pair of circles x^{2} + y^{2} – 4x – 6y – 12 = 0, x^{2} + y^{2} + 6x + 18y + 26 = 0.

Solution:

Given equations of the circles are

x^{2} + y^{2} – 4x – 6y – 12 = 0 …….(1)

x^{2} + y^{2} + 6x + 18y + 26 = 0 ………(2)

for the circle (1), centre c_{1} = (2, 3)

Now, r_{1} + r_{2} = 5 + 8 = 13

∴ c_{1}c_{2} = r_{1} + r_{2}

∴ The given circles touch each other externally.

Question 111.

Discuss the relative position of the pair of circles x^{2} + y^{2} – 2x + 4y – 4 = 0, x^{2} + y^{2} + 4x – 6y – 3 = 0.

Solution:

Given equations of the circles are

x^{2} + y^{2} – 2x + 4y – 4 = 0 …….(1)

x^{2} + y^{2} + 4x – 6y – 3 = 0 ………..(2)

For the circle (1), centre c_{1} = (1, -2)

Question 112.

Find the number of possible common tangents that exist for the pairs of circles x^{2} + y^{2} + 6x + 6y + 14 = 0, x^{2} + y^{2} – 2x – 4y – 4 = 0.

Solution:

Given equations of the circles are

x^{2} + y^{2} + 6x + 6y + 14 = 0 ……..(1)

x^{2} + y^{2} – 2x – 4y – 4 = 0 ……….(2)

For the circle (1), Centre c_{1} = (-3, -3)

Now, r_{1} + r_{2} = 2 + 3 = 5 = √25

∴ c_{1}c_{2} > r_{1} + r_{2}

∴ Given circles, each circle lies completely outside the other.

∴ The number of common tangents = 4.

Question 113.

Find the number of possible common tangents that exist for the pairs of circles x^{2} + y^{2} – 4x – 2y + 1 = 0, x^{2} + y^{2} – 6x – 4y + 4 = 0.

Solution:

Given equations of the circles are

x^{2} + y^{2} – 4x – 2y + 1 = 0 ……..(1)

x^{2} + y^{2} – 6x – 4y + 4 = 0 ……….(2)

For the circle (1), Centre c_{1} = (2, 1)

Now, |r_{1} – r_{2}| = |2 – 3| = |-1| = 1

r_{1} + r_{2} = 2 + 3 = 5 = √25

∴ |r_{1} – r_{2}| < c_{1}c_{2} < r_{1} + r_{2}

∴ The given circles intersect at two points.

∴ The number of common tangents = 2.

Question 114.

Find the number of possible common tangents that exist for the pairs of circles x^{2} + y^{2} – 4x + 2y – 4 = 0, x^{2} + y^{2} + 2x – 6y + 6 = 0.

Solution:

Given equations of the circles are

x^{2} + y^{2} – 4x + 2y – 4 = 0 ……..(1)

x^{2} + y^{2} + 2x – 6y + 6 = 0 ……..(2)

For the circle (1), Centre c_{1} = (2, -1)

Now, r_{1} + r_{2} = 3 + 2 = 5

∴ c_{1}c_{2} = r_{1} + r_{2}

∴ The given circles touch each other externally.

∴ The number of common tangents = 3.

Question 115.

Find the number of possible common tangents that exist for the pairs of circles x^{2} + y^{2} + 4x – 6y – 3 = 0, x^{2} + y^{2} + 4x – 2y + 4 = 0.

Solution:

Given equations of the circles are

x^{2} + y^{2} + 4x – 6y – 3 = 0 ……..(1)

x^{2} + y^{2} + 4x – 2y + 4 = 0 ……….(2)

For the circle (1), Centre c_{1} = (-2, 3)

Now, |r_{1} + r_{2}| = |4 + 1| = 5

∴ c_{1}c_{2} < |r_{1} + r_{2}|

∴ In the given circles one circle lies completely inside the other.

∴ The number of common tangents = 0.

Question 116.

Find the equation of the pair of tangents from (10, 4) to the circle x^{2} + y^{2} = 25.

Solution:

The given equation of the circle is x^{2} + y^{2} = 25.

Here g = 0, f = 0, c = -25

Let, the given point P(x_{1}, y_{1}) = (10, 4)

∴ The equation of the pair of tangents drawn from (10, 4) to the circle x^{2} + y^{2} = 25 is

SS_{11} = \(\mathrm{S}_1{ }^2\)

⇒ (x^{2} + y^{2} + 2gx + 2fy+ c) \(\left(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{fy}_1+\mathrm{c}\right)\) = (xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c)^{2}

⇒ (x^{2} + y^{2} – 25) [(10)^{2} + (4)^{2} + 2(0)(10) + 2(0)(4) – 25] = [x(10) + y(4) + 0(x + 10) + 0(y + 4) – 25]^{2}

⇒ (x^{2} + y^{2} – 25) (100 + 16 – 25) = (10x + 4y – 25)^{2}

⇒ (x^{2} + y^{2} – 25) (91) = (10x + 4y – 25)^{2}

⇒ 91x^{2} + 91y^{2} – 2275 = 100x^{2} + 16y^{2} + 625 + 80xy – 200y – 500x

⇒ 9x^{2} + 80xy – 75y^{2} – 500x – 200y + 2900 = 0