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

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

Question 1.

If a point P is moving such that the lengths of tangents drawn from P to the circles x^{2} + y^{2} – 4x – 6y – 12 = 0 and x^{2} + y^{2} + 6x + 18y + 26 = 0 are in the ratio 2 : 3, then find the equation of the locus of P. [(AP) Mar. ’19, (TS) ’17]

Solution:

Let P(x_{1}, y_{1}) be a point on the locus and S = x^{2} + y^{2} – 4x – 6y – 12 = 0

S’ = x^{2} + y^{2} + 6x + 18y + 26 = 0 be the given circles.

Question 2.

If a point P is moving such that the lengths of tangents drawn from P to x^{2} + y^{2} – 2x + 4y – 20 = 0 and x^{2} + y^{2} – 2x – 8y + 1 = 0 are in the ratio 2 : 1, then show that the equation of locus of P is x^{2} + y^{2} – 2x – 12y + 8 = 0.

Solution:

Let P(x, y) be a point on the locus and

S = x^{2} + y^{2} – 2x + 4y – 20 = 0

S’ = x^{2} + y^{2} – 2x – 8y + 1 = 0 be the given circles.

Length of tangent from P to S = 0 is

PA = \(\sqrt{S_{11}}=\sqrt{x^2+y^2-2 x+4 y-20}\)

Length of tangent from P to S’ = 0 is

PB = \(\sqrt{\mathrm{S}_{11}^{\prime}}=\sqrt{\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-8 \mathrm{y}+1}\)

Given condition is PA : PB = 2 : 1

\(\frac{\mathrm{PA}}{\mathrm{PB}}=\frac{2}{1}\)

⇒ PA = 2PB

⇒ \(\sqrt{x^2+y^2-2 x+4 y-20}\) = \(2 \sqrt{x^2+y^2-2 x-8 y+1}\)

squaring on both sides

⇒ x^{2} + y^{2} – 2x + 4y – 20 = 4(x^{2} + y^{2} – 2x – 8y + 1)

⇒ x^{2} + y^{2} – 2x + 4y – 20 – 4x^{2} – 4y^{2} + 8x + 32y – 4 = 0

⇒ -3x^{2} – 3y^{2} + 6x + 36y – 24 = 0

⇒ x^{2} + y^{2} – 2x – 12y + 8 = 0

The equation of locus of P is x^{2} + y^{2} – 2x – 12y + 8 = 0.

Question 3.

Find the equations of the tangent to the circle x^{2} + y^{2} – 4x + 6y – 12 = 0 which are parallel to x + y – 8 = 0. (Mar. ’01)

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

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

Radius r = \(\sqrt{g^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{4+9+12}\) = 5

Given equation of the straight line is x + y – 8 = 0

The equation of the tangent parallel to x + y – 8 = 0 is

x + y + k = 0 ……….(1)

Since eq. (1) is a tangent to the given circle then r = d.

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

⇒ 5 = \(\frac{|1(2)+1(-3)+\mathbf{k}|}{\sqrt{(1)^2+(1)^2}}\)

⇒ 5 = \(\frac{|2-3+\mathbf{k}|}{\sqrt{2}}\)

⇒ 5√2 = |k – 1|

⇒ k – 1 = ±5√2

⇒ k = 1 ± 5√2

Substitute the value of ‘k’ in eq. (1)

x + y + 1 ± 5√2 = 0

Question 4.

Show that x + y + 1 = 0 touches the circle x^{2} + y^{2} – 3x + 7y + 14 = 0 and find its point of contact.

Solution:

Given equation of the circle is x^{2} + y^{2} – 3x + 7y + 14 = 0

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

Given the equation of the line is x + y + 1 = 0.

Now, d = The perpendicular distance from the centre C\(\left(\frac{3}{2}, \frac{-7}{2}\right)\) to the line x + y + 1 = 0.

Since r = d then, the line x + y + 1 = 0 touches the circle x^{2} + y^{2} – 3x + 7y + 14 = 0.

Let P(h, k) be the point of contact.

Now, P(h, k) is the foot of the perpendicular drawn from C = \(\left(\frac{3}{2}, \frac{-7}{2}\right)\) to the line x + y + 1 = 0

∴ The point of contact is P(2, -3).

Question 5.

Show that the points (1, 1), (-6, 0), (-2, 2), and (-2, -8) are concyclic and find the equation of the circle on which they lie. [(AP) Mar. ’19; May ’17]

Solution:

Let, the equation of the required circle is

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

Since, (1) passes through point (1, 1)

1^{2} + 1^{2} + 2g(1) + 2f(1) + c = 0

2 + 2g + 2f + c = 0

2g + 2f + c = -2 ……(2)

Since, (1) passes through the point (-6, 0)

(-6)^{2} + 0^{2} + 2g(-6) + 0 + c = 0

36 + c – 12g = 0

-12g + c = -36 …….(3)

Since, (1) passes through the point (-2, 2)

(-2)^{2} + (2)^{2} + 2g(-2) + 2f(2) + c = 0

4 + 4 – 4g + 4f + c = 0

-4g + 4f + c = -8 ……..(4)

From (2) and (3)

From (3) and (4)

Solving (5) and (6)

g = 2, f = 3

Substitute the values of g, f in (2)

2(2) + 2(3) + c = -2

4 + 6 + c = -2

10 + c = -2

c = -12

Now, substitute the values of g, f, c in (1)

∴ The equation of the required circle is

x^{2} + y^{2} + 2(2) x + 2(3) y – 12 = 0

x^{2} + y^{2} + 4x + 6y – 12 = 0 ………(7)

Now, substituting the point (-2, -8) in (7)

(-2)^{2} + (-8)^{2} + 4(-2) + 6(-8) – 12 = 0

4 + 64 – 8 – 48 – 12 = 0

68 – 68 = 0

0 = 0

∴ Given points are concyclic.

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

Question 6.

Find the equations of the tangents to the circle x^{2} + y^{2} + 2x – 2y – 3 = 0 which are perpendicular to 3x – y + 4 = 0.

Solution:

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

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

we get g = 1, f = -1, c = -3,

Centre of the circle C(-g, -f) = (-1, 1)

The radius of the circle

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

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

= √5

Given the equation of the straight line is 3x – y + 4 = 0

The equation of the straight line perpendicular to 3x – y + 4 = 0 is

x + 3y + k = 0 ……..(1)

Since (1) is the tangent to the given circle then r = d

√5 = \(\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}\)

The equations of the tangents are from (1),

x + 3y – 2 ± 5√2 = 0

Question 7.

Show that the tangent at (-1, 2) of the circle x^{2} + y^{2} – 4x – 8y + 7 = 0 touches the circle x^{2} + y^{2} + 4x + 6y = 0 and finds its point of contact. (May ’10)

Solution:

Given equation of the circle is x^{2} + y^{2} – 4x – 8y + 7 = 0

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

we get g = -2, f = -4, c = 7

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

The equation of the tangent is S1 = 0

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

⇒ x(-1) + y(2) – 2(x – 1) – 4(y + 2) + 7 = 0

⇒ -x + 2y – 2x + 2 – 4y – 8 + 7 = 0

⇒ -3x – 2y + 1 = 0

⇒ 3x + 2y – 1 = 0

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

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

we get g = 2, f = 3, c = 0

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

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

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

= √13

Now d = The perpendicular distance from the centre C = (-2, -3) to the line 3x + 2y – 1 = 0

Since r = d, the tangent at (-1, 2) of the circle x^{2} + y^{2} – 4x – 8y + 7 = 0 touches the circle x^{2} + y^{2} + 4x + 6y = 0.

Let Q(h, k) be the point of contact.

Now, Q(h, k) is the foot of the perpendicular drawn from centre C(-2, -3) to the line 3x + 2y – 1 = 0.

∴ The point of contact Q(h, k) = (1, -1).

Question 8.

Find the angle between the tangents drawn from (3, 2) to the circle x^{2} + y^{2} – 6x + 4y – 2 = 0. (Mar. ’12)

Solution:

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

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

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

Radius r = \(\sqrt{9+4+2}=\sqrt{15}\)

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

Length of the tangent = \(\sqrt{\mathrm{S}_{11}}\)

Question 9.

Find the locus of ‘P’ where the tangent is drawn from ‘P’ to x^{2} + y^{2} = a^{2} are perpendicular to each other.

Solution:

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

Radius r = a

Let P(x_{1}, y_{1}) be any point on the locus.

Length of the tangent = \(\sqrt{\mathrm{S}_{11}}\) = \(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2-\mathrm{a}^2}\)

Given that angle between the tangents θ = 90°

If ‘θ’ is the angle between the tangents through ‘P’ to the given circle then

\(\tan \left(\frac{\theta}{2}\right)=\frac{r}{\sqrt{S_{11}}}\)

Question 10.

Find the chord length intercepted by the circle x^{2} + y^{2} – 8x – 2y – 8 = 0 on the line x + y + 1 = 0. [(TS) Mar. ’16]

Solution:

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

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

we get g = -4, f = -1, c = -8

Centre of the circle C = (-g, -f) = (4, 1)

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

= \(\sqrt{(-4)^2+(-1)^2+8}\)

= 5

Given equation of the straight line is x + y + 1 = 0

Now d = perpendicular distance from the centre C(4, 1) to the chord x + y + 1 = 0

Question 11.

Find the chord length intercepted by the circle x^{2} + y^{2} – x + 3y – 22 = 0 on the line y = x – 3. [(TS) Mar. ’20; (AP) Mar. ’18, May ’16, Mar. ’13]

Solution:

Given equation of the circle is x^{2} + y^{2} – x + 3y – 22 = 0

Question 12.

Find die length of the chord formed by x^{2} + y^{2} = a^{2} on the line x cos α + y sin α = P. [(TS) Mar. ’16]

Solution:

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

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

We get g = 0, f = 0, c = -a^{2}

Centre of the circle C(-g, -f) = (0, 0)

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

= \(\sqrt{(0)^2+(0)^2+a^2}\)

= a

Given the equation of the straight line is x cos α + y sin α – P = 0

Now d = perpendicular distance from the centre C(0,0) to the chord x cos α + y sin α – P = 0

Question 13.

Find the equation of the circle with centre (-2, 3) cutting a chord length 2 units on 3x + 4y + 4 = 0. (Mar. ’11)

Solution:

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

Given the equation of the straight line is 3x + 4y + 4 = 0

Now d = The perpendicular distance from the centre C(-2, 3) to the line 3x + 4y + 4 = 0.

Given that the length of the chord = 2

\(2 \sqrt{\mathrm{r}^2-\mathrm{d}^2}\) = 2

⇒ \(\sqrt{\mathrm{r}^2-\mathrm{d}^2}\) = 1

⇒ r^{2} – d^{2} = 1

⇒ r^{2} – 2^{2} = 1

⇒ r^{2} – 4 = 1

⇒ r^{2} = 5

⇒ r = √5

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

⇒ (x + 2)^{2} + (y- 3)^{2} = (√5)^{2}

⇒ x^{2} + 4 + 4x + y^{2} + 9 – 6y = 5

⇒ x^{2} + y^{2} + 4x – 6y + 8 = 0

Question 14.

Find the area of the triangle formed by the normal at (3, -4) to the circle x^{2} + y^{2} – 22x – 4y + 25 = 0 with the coordinate axis. [Mar. ’18 (TS)]

Solution:

Given circle is x^{2} + y^{2} – 22x – 4y + 25 = 0

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

we get 2g = -22 ⇒ g = -11

2f = -4 ⇒ f = -2, c = 25

centre c(-g, -f) = c(11, 2)

The area of the triangle formed by the normal with the coordinate axis = \(\frac{c^2}{2|a b|}\)

= \(\frac{(-25)^2}{2|3(-4)|}\)

= \(\frac{625}{24}\) sq units

Question 15.

Find the mid point of the chord intercepted by x^{2} + y^{2} – 2x – 10y + 1 = 0 on the line x – 2y + 7 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 2x – 10y + 1 = 0

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

we get, g = -1, f = -5, c = 1

Centre of the circle, C = (-g, -f) = (1, 5)

Given equation of the straight line is x – 2y + 7 = 0

Comparing this equation with ax + by + c = 0,

we get a = 1, b = -2, c = 7

Let P(h, k) is the midpoint of the chord x – 2y + 7 = 0

Now P(h, k) is the foot of the perpendicular from centre C(1, 5) on the chord x – 2y + 7 = 0

Question 16.

Show that the area of the triangle formed by the two tangents through P(x_{1}, y_{1}) to the circle S = x^{2} + y^{2} + 2gx + 2fy + c = 0 and the chord of contact of ‘P’ with respect to S = 0 is \(\frac{\mathbf{r}\left(\mathbf{S}_{11}\right)^{3 / 2}}{\mathbf{S}_{11}+\mathbf{r}^2}\), where ‘r’ is the radius of the circle.

Solution:

If θ is the angle between the tangents ‘P’ to S = 0, then

\(\tan \left(\frac{\theta}{2}\right)=\frac{r}{\sqrt{S_{11}}}\)

Let Q, R be the chord of the contact from ‘P’ to the circle.

Let PA be the ⊥r from P to QR.

Question 17.

Find the pole of the line x + y + 2 = 0 w.r.t the circle x^{2} + y^{2} – 4x + 6y – 12 = 0. [(AP) Mar. ’17, May ’15]

Solution:

Given equation of the circle 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

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

= \(\sqrt{4+9+12}\)

= 5

Given equation of the straight line is x + y + 2 = 0

Comparing this equation with lx + my + n = 0,

we get l = 1, m = 1, n = 2

The pole of lx + my + n = 0 w.r.t x^{2} + y^{2} + 2gx + 2fy + c = 0 is

Question 18.

Find the pole of the line 3x + 4y – 45 = 0 w.r.t the circle x^{2} + y^{2} – 6x – 8y + 5 = 0. [(AP) Mar. ’16]

Solution:

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

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

We get g = -3, f = -4, c = 5

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

= \(\sqrt{9+16-5}\)

= √20

Given the equation of the straight line is 3x + 4y – 45 = 0

Comparing the given equation with lx + my + n = 0,

we get l = 3, m = 4, n = -45

The pole of lx + my + n = 0 w.r.t x^{2} + y^{2} + 2gx + 2fy + c = 0 is

Question 19.

Show that the lines 2x + 3y + 11 = 0, 2x – 2y – 1 = 0 are conjugate w.r.t the circle x^{2} + y^{2} + 4x + 6y + 12 = 0.

Solution:

Given equation of the circle 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

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

= \(\sqrt{4+9-12}\)

= 1

Given equations of the straight lines are

2x + 3y + 11 = 0 …….(1)

2x – 2y – 1 = 0 ………(2)

Comparing (1) with l_{1}x + m_{1}y + n_{1} = 0

we get l_{1} = 2, m_{1} = 3, n_{1} = 11

Comparing (2) with l_{2}x + m_{2}y + n_{2} = 0

we get l_{2} = 2, m_{2} = -2, n_{2} = -1

Now (l_{1}g + m_{1}f – n_{1}) (l_{2}g + m_{2}f – n_{2})

= [2(2) + 3(3) – 11] [2(2) + 3(-2) + 1]

= (4 + 9 – 11) (4 – 6 + 1)

= 2(-1)

= -2

r^{2} (l_{1}l_{2} + m_{1}m_{2}) = (1)^{2} [2(2) + 3(-2)]

= 1(4 – 6)

= -2

∴ (l_{1}g + m_{1}f – n_{1}) (l_{2}g + m_{2}f – n_{2}) = r^{2} (l_{1}l_{2} + m_{1}m_{2})

∴ Given lines are conjugate w.r.t given circle.

Question 20.

Find the value of ‘k’ if x + y – 5 = 0 and 2x + ky – 8 = 0 are conjugate with respect to the circle x^{2} + y^{2} – 2x – 2y – 1 = 0. [(TS) May ’18]

Solution:

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

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

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

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

= \(\sqrt{1+1+1}\)

= √3

Given equations of the straight lines are

x + y – 5 = 0 ………(1)

2x + ky – 8 = 0 ……..(2)

Comparing (1) with l_{1}x + m_{1}y + n_{1} = 0

we get l_{1} = 1, m_{1} = 1, n_{1} = -5

Comparing (2) with l_{2}x + m_{2}y + n_{2} = 0

we get l_{2} = 2, m_{2} = k, n_{2} = -8

Since given lines are conjugate w.r.t given circle, then

(l_{1}g + m_{1}f – n_{1}) (l_{2}g + m_{2}f – n_{2}) = r^{2} (l_{1}l_{2} + m_{1}m_{2})

⇒ [1(-1) + 1(-1) + 5] [2(-1) + k(-1) + 8] = (√3)2 [1(2) + 1(k)]

⇒ (-1 – 1 + 5)(-2 – k + 8) = 3(2 + k)

⇒ 3(-k + 6) = 3(k + 2)

⇒ -k + 6 = k + 2

⇒ 2k = 4

⇒ k = 2

Question 21.

Find the condition that the tangents drawn from (0, 0) to S = x^{2} + y^{2} + 2gx + 2fy + c = 0 be perpendicular to each other. [(TS) May ’16]

Solution:

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

Centre, C = (-g, -f)

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

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

The angle between the tangents, θ = 90°

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

Question 22.

If the abscissae of points A, B are the roots of the equation x^{2} + 2ax – b^{2} = 0 and ordinates of A, B are roots of y^{2} + 2py – q^{2} = 0, then find the equation of a circle for which \(\overline{\mathbf{A B}}\) is a diameter. (Mar. ’14)

Solution:

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

Since x_{1}, x_{2} are the roots of the quadratic equation x^{2} + 2ax – b^{2} = 0 then

sum of the roots = \(\frac{-b}{a}\)

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

Product of the roots = \(\frac{c}{a}\)

x_{1}x_{2} = \(\frac{-b^2}{1}\) = -b^{2}

Since y_{1}, y_{2} are the roots of the quadratic equation y^{2} + 2py – q^{2} = 0 then

sum of the roots = \(\frac{-b}{a}\)

y_{1} + y_{2} = \(\frac{-2 p}{1}\) = -2p

Product of the roots = \(\frac{c}{a}\)

y_{1}y_{2} = \(\frac{-\mathrm{q}^2}{1}\) = -q^{2}

The equation of a circle for which \(\overline{\mathbf{A B}}\) is a diameter is

(x – x_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0

⇒ x^{2} – xx_{2} – xx_{1} + x_{1}x_{2} + y^{2} – yy_{1} – yy_{2} + y_{1}y_{2} = 0

⇒ x^{2} – x(x_{1} + x_{2}) + x_{1}x_{2} + y^{2} – y(y_{1} + y_{2}) + y_{1}y_{2} = 0

⇒ x^{2} – x(-2a) + (-b2) + y^{2} – y(-2p) + (-q2) = 0

⇒ x^{2} + y^{2} + 2ax + 2py – b^{2} – q^{2} = 0

Question 23.

Find the equation of the circle which touches the x-axis at a distance of 3 from the origin and make an intercept of length 6 on the y-axis.

Solution:

Let the equation of the required circle is

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

Since (1) meets the x-axis at A(3, 0)

∴ Point A(3, 0) lies on the circle (1), then

(3)^{2} + (0)^{2} + 2g(3) + 2f(0) + c = 0

⇒ 9 + 6g + c = 0

⇒ 6g + c = -9 ……..(2)

Since, the circle (1) touches the x-axis, then

g^{2} = c …….(3)

From (2) and (3)

6g + g^{2} = – 9

⇒ g^{2} + 6g + 9 = 0

⇒ (g + 3)^{2} = 0

⇒ g + 3 = 0

⇒ g = -3

Now, substitute the value of g in (3), and we get

(-3)^{2} = c

⇒ c = 9

Given that the intercept on the y-axis made by (1) is 6

\(2 \sqrt{f^2-c}\) = 6

⇒ \(\sqrt{\mathrm{f}^2-\mathrm{c}}\) = 3

⇒ \(\sqrt{\mathrm{f}^2-9}\) = 3

⇒ f^{2} – 9 = 9

⇒ f^{2} = 18

⇒ f = ±3√2

Substitute the values of g, f, c in (1)

∴ The required equation of the circle is x^{2} + y^{2} + 2(-3)x + 2(±3√2)y + 9 = 0

⇒ x^{2} + y^{2} – 6x ± 6√2y + 9 = 0

Question 24.

Find the equation of the circle passing through (0, 0) and making intercepts 4, 3 on the x, y-axis respectively.

Solution:

Let the equation of the required circle is

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

Since (1) passes through the point (0, 0), then

(0)^{2} + (0)^{2} + 2g(0) + 2f(0) + c = 0

∴ c = 0

Given that the intercept on the x-axis made by (1) = 4

Now, substitute the values of g, f, c in (1)

The equation of the required circle is

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

⇒ x^{2} + y^{2} ± 4x ± 3y = 0

Question 25.

Show that the locus of the point of intersection of the lines x cos α + y sin α = a, x sin α – y cos α = b (α is a parameter) is a circle.

Solution:

Given equations of the straight lines are

x cos α + y sin α = a …….(1)

x sin α – y cos α = b …….(2)

Now (1)^{2} + (2)^{2}

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

⇒ x^{2} cos^{2}α + y^{2} sin^{2}α + 2xy sin α cos α + x^{2} sin^{2}α + y^{2} cos^{2}α – 2xy sin α cos α = a^{2} + b^{2}

⇒ x^{2} (cos^{2}α + sin^{2}α) + y^{2} (sin^{2}α + cos^{2}α) = a^{2} + b^{2}

⇒ x^{2} (1) + y^{2} (1) = a^{2} + b^{2}

∴ x^{2} + y^{2} = a^{2} + b^{2}

∴ The locus of the point of intersection of the given line is x^{2} + y^{2} = a^{2} + b^{2}

It represents a circle.

Question 26.

If y = mx + c and x^{2} + y^{2} = a^{2}

(i) Intersect at A and B

(ii) AB = 2λ, then show that c^{2} = (1 + m^{2}) (a^{2} – λ^{2})

Solution:

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

Comparing this equation with x^{2} + y^{2} = r^{2}, then centre of the circle C = (0, 0)

The radius of the circle r = a

Given the equation of the straight line is mx – y + c = 0

Now, d = the perpendicular distance from the centre C(0, 0) to the line mx – y + c = 0

Question 27.

Find the equation of the circle with centre (2, 3) and touch the line 3x – 4y + 1 = 0.

Solution:

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

Given the equation of the straight line is 3x – 4y + 1 = 0

Since, the line 3x – 4y + 1 = 0 touches the required circle, then the line 3x – 4y + 1 = 0 is a tangent to the required circle.

∴ Radius r = The perpendicular distance from the centre C(2, 3) to the tangent 3x – 4y + 1 = 0

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

⇒ (x – 2)^{2} + (y – 3)^{2} = (1)^{2}

⇒ x^{2} + 4 – 4x + y^{2} + 9 – 6y = 1

⇒ x^{2} + y^{2} – 4x – 6y + 12 = 0

Question 28.

Find the equation of the tangent at the point 30° (parametric value of θ) of the circle x^{2} + y^{2} + 4x + 6y – 39 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} + 4x + 6y – 39 = 0

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

We get g = 2, f = 3, c = -39

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

= \(\sqrt{(2)^2+(3)^2+39}\)

= 2√13

The given point θ = 30°

∴ The equation of the tangent at the point θ of the given circle is (x + g) cos θ + (y + f) sin θ = r

⇒ (x + 2) cos 30° + (y + 3) sin 30° = 2√13

⇒ (x + 2) . \(\frac{\sqrt{3}}{2}\) + (y + 3) . \(\frac{1}{2}\) = 2√13

⇒ √3x + 2√3 + y + 3 = 4√13

⇒ √3x + y + 2√3 + 3 – 4√13 = 0

Question 29.

Find the equation of the tangent to x^{2} + y^{2} – 2x + 4y = 0 at (3, -1). Also, find the equation of tangent parallel to it. [(TS) May ’17]

Solution:

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

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

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

The given point P(x_{1}, y_{1}) = (3, -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(3) + y(-1) – 1(x + 3) + 2(y – 1) + 0 = 0

⇒ 3x – y – x – 3 + 2y – 2 = 0

⇒ 2x + y – 5 = 0

Centre of the circle, C = (-g, -f) = (1, -2)

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

= \(\sqrt{(-1)^2+(2)^2+0}\)

= √5

The equation of the straight line parallel to the tangent 2x + y – 5 = 0 is

2x + y + k = 0 …….(1)

If (1) is a tangent to the given circle when r = d

The equations of tangents to the circle are from (1)

2x + y ± 5 = 0

One of these equations namely 2x + y – 5 = 0 is the tangent at (3, -1).

The tangent parallel to 2x + y – 5 = 0 is 2x + y + 5 = 0.

Question 30.

If 4x – 3y + 7 = 0 is a tangent to the circle represented by x^{2} + y^{2} – 6x + 4y – 12 = 0 then find the point of contact.

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

Centre of the circle, C = (-g, -f) = (3, -2)

Given equation of the tangent is 4x – 3y + 7 = 0

Let the point of contact be P = (h, k)

Now, P(h, k) is the foot of the perpendicular from the centre, C(3, -2) to the tangent 4x – 3y + 7 = 0

Question 31.

Find the equation of the normal to the circle x^{2} + y^{2} – 4x – 6y + 11 = 0 at (3, 2). Also, find the other point where the normal meets the circle.

Solution:

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

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

Centre of the circle C(-g, -f) = (2, 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(3) + y(2) + (- 2) (x + 3) – 3(y + 2) + 11 = 0

⇒ 3x + 2y – 2x – 6 – 3y – 6 + 11 = 0

⇒ x – y – 1 = 0

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

Slope of the normal at P is \(\frac{-1}{m}=\frac{-1}{1}\) = -1

∴ The equation of the normal at P is

y – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

⇒ y – 2 = -1(x – 3)

⇒ y – 2 = -x + 3

⇒ x + y – 5 = 0

Other point of the normal Q = (x, y)

The centre of the circle is the midpoint of P and Q (point of intersection of normal and circle)

∴ C(2, 3) = Midpoint of PQ

∴ The normal at (3, 2) meets the circle at (1, 4).

Question 32.

Find the equation of the normal to the circle x^{2} + y^{2} – 10x – 2y + 6 = 0 at (3, 5). Also, find the other point where the normal meets the circle.

Solution:

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

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

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

\(\mathrm{y}-\mathrm{y}_1=\frac{-1}{\mathrm{~m}}\left(\mathrm{x}-\mathrm{x}_1\right)\)

⇒ y – 5 = -2(x – 3)

⇒ y – 5 = -2x + 6

⇒ 2x + y – 1 = 0

Question 33.

Find the locus of P, where the tangents drawn from P to x^{2} + y^{2} = a^{2} include an angled α.

Solution:

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

Comparing this equation with x^{2} + y^{2} = r^{2}

We get r = a

Let P(x_{1}, y_{1}) be a point on the locus.

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

Question 34.

If ax + by + c = 0 is polar of (1, 1) w.r.t x^{2} + y^{2} – 2x + 2y + 1 = 0 and HCF of a, b, c is equal to one, then find a^{2} + b^{2} + c^{2}.

Solution:

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

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

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

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

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

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

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

⇒ x + y – x – 1 + y + 1 + 1 = 0

⇒ 2y + 1 = 0

Given equation of the polar is ax + by + c = 0, then

a = 0, b = 2, c = 1

The H.C.F of a, b, c is equal to ‘1’.

Now a^{2} + b^{2} + c^{2} = (0)^{2} + (2)^{2} + (1)^{2}

= 0 + 4 + 1

= 5

Question 35.

If the polar of the points on the circle x^{2} + y^{2} = a^{2} with respect to the circle x^{2} + y^{2} = b^{2} touches the circle x^{2} + y^{2} = c^{2}, then prove that a, b, c are in Geometric Progression.

Solution:

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

Let P(x_{1}, y_{1}) be any point on the circle x^{2} + y^{2} = a^{2}, then

\(\mathrm{x}_1^2+\mathrm{y}_1^2=\mathrm{a}^2\) ……….(1)

The equation of the second circle is x^{2} + y^{2} = b^{2}

The polar of P(x_{1}, y_{1}) w.r.t the second circle x^{2} + y^{2} = b^{2} is S_{1} = 0

xx_{1} + yy_{1} – b^{2} = 0 ……..(2)

The equation of the third circle is x^{2} + y^{2} = c^{2}

Centre C = (0, 0); Radius r = c

Since (2) is a tangent to the circle x^{2} + y^{2} = c^{2} then r = d

c = \(\frac{\left|\mathrm{x}_1(0)+\mathrm{y}_1(0)-\mathrm{b}^2\right|}{\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}}\)

⇒ c = \(\frac{\left|-b^2\right|}{\sqrt{a^2}}\)

⇒ c = \(\frac{\mathrm{b}^2}{\mathrm{a}}\)

⇒ b^{2} = ac

∴ a, b, c are in geometric progression.

Question 36.

Find the slope of the polar of (1, 3) with respect to the circle x^{2} + y^{2} – 4x – 4y – 4 = 0. Also, find the distance from the centre of it.

Solution:

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

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

We get g = -2, f = -2, c = -4

Centre of the circle C = (-g, -f) = (2, 2)

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

The equation of polar of (1, 3) w.r.t the given circle is S1 = 0

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

⇒ x(1) + y(3) – 2(x + 1) – 2(y + 3) – 4 = 0

⇒ x + 3y – 2x – 2 – 2y – 6 – 4 = 0

⇒ -x + y – 12 = 0

⇒ x – y + 12 = 0

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

The perpendicular distance from the centre C(2, 2) to the polar x – y + 12 = 0 is

Question 37.

Show that, four common tangents can be drawn for the circles given by x^{2} + y^{2} – 14x + 6y + 33 = 0 and x^{2} + y^{2} + 30x – 2y + 1 = 0 and find the internal and external centres of similitude. [(TS) Mar. ’19]

Solution:

Given equations of the circles are

x^{2} + y^{2} – 14x + 6y + 33 = 0 ……..(1)

x^{2} + y^{2} + 30x – 2y + 1 = 0 ………(2)

For the circle (1), Centre C_{1} = (7, -3)

Now r_{1} + r_{2} = 5 + 15 = 20 = √400

∴ C_{1}C_{2} > r_{1} + r_{2}

∴ The given circles are each circle lies completely outside the other circle.

∴ No.of common tangents = 4

The internal centre of similitude A_{1} internally divides C_{1}C_{2} in the ratio r_{1} : r_{2} (5 : 15 = 1 : 3).

∴ The internal centre of similitude

The external centre of similitude A_{2} divides C_{1}C_{2} in the ratio r_{1} : r_{2} (5 : 15 = 1 : 3) externally.

∴ The external centre of similitude

Question 38.

If a point P is moving such that the lengths of the tangent drawn from ‘P’ to the circle x^{2} + y^{2} + 8x + 12y + 15 = 0 and x^{2} + y^{2} – 4x – 6y – 12 = 0 are equal, then find the equation of the locus of P. (Mar. ’09)

Solution:

Given equations of the circles are

x^{2} + y^{2} + 8x + 12y + 15 = 0

x^{2} + y^{2} – 4x – 6y – 12 = 0

Let P(x_{1}, y_{1}) be any point on the locus.

PA, PB be the lengths of the tangent from P to the circles (1) & (2) respectively.

Given condition is PA = PB

The equation of the locus of ‘P’ is 12x + 18y + 27 = 0

⇒ 4x + 6y + 9 = 0

Question 39.

Find the inverse point of (-2, 3) with respect to the circle x^{2} + y^{2} – 4x – 6y + 9 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 4x – 6y + 9 = 0

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

we get, g = -2, f = -3, c = 9

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

Now, the polar of 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) – 2(x – 2) – 3(y + 3) + 9 = 0

⇒ -2x + 3y – 2x + 4 – 3y – 9 + 9 = 0

⇒ -4x + 4 = 0

⇒ x – 1 = 0

Let Q(h, k) be the inverse point of P(-2, 3) w.r.t. the circle x^{2} + y^{2} – 4x – 6y + 9 = 0.

Now, Q(h, k) is the foot of the perpendicular from P(-2, 3) w.r.t. x – 1 = 0

∴ The inverse point of (-2, 3) is (1, 3).