Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Circle Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2B Circle Important Questions

Question 1.

If x^{2} + y^{2} + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3), find g, f and its radius. [Mar. 11]

Solution:

Circle is x^{2} + y^{2} + 2gx + 2fy – 12 = 0

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

∴ g = – 2, f = – 3, c = – 12

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

= \(\sqrt{4+9+12}\) = 5 units

Question 2.

Obtain the parametric equation of x^{2} + y^{2} = 4 [Mar. 14]

Solution:

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

C (0, 0), r = 2

Parametric equations are

x = – g + r cos θ = 2 cos θ

y = – b + r sin θ = 2 sin θ, 0 < θ < 2π

Question 3.

Obtain the parametric equation of (x – 3)^{2} + (y – 4)^{2} = 8^{2} [A.P. Mar. 16; Mar. 11]

Solution:

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

Centre (3, 4), r = 8

Parametric equations are

x = 3 + 8 cos θ, y = 4 + 8 sin θ, 0 < θ < 2π

Question 4.

Find the power of the point P with respect to the circle S = 0 when

ii) P = (-1,1) and S ≡ x^{2} + y^{2} – 6x + 4y – 12

Solution:

Power of the point = S_{11}

= 1 + 1 + 6 + 4 – 12 = 0

iii) P = (2, 3) and S ≡ x^{2} + y^{2} – 2x + 8y – 23

Power of the point = S_{11}

= 4 + 9 – 4 + 24 – 23 = 10

iv) P = (2, 4) and S ≡ x^{2} + y^{2} – 4x – 6y – 12

Power of the point = 4 + 16 – 8 – 24 – 12

= -24.

Question 5.

If the length of the tangent from (5, 4) to the circle x^{2} + y^{2} + 2ky = 0 is 1 then find k. [Mar. 15; Mar. 01]

Solution:

= \(\sqrt{S_{11}}=\sqrt{(5)^{2}+(4)^{2}+8 k}\)

But length of tangent = 1

∴ 1 = \(\sqrt{25+16+8k}\)

Squaring both sides we get 1 = 41 + 8k

k = – 5 units.

Question 6.

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

Solution:

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

Equation of the polar is S_{1} = 0

Polar of P(1,-2) is

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 7.

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. [T.S. Mar. 16]

Solution:

Equation of the circle is

x^{2} + y^{2} = 35

Polar of P(1, 3) is x. 1 + y. 3 = 35

x + 3y = 35

P(1, 3) and Q(2, k) are conjugate points

The polar of P passes through Q

2 + 3k = 35

3k = 33

k = 11

Question 8.

If the circle x^{2} + y^{2} – 4x + 6y + a = 0 has radius 4 then find a. [A.P. Mar. 16]

Solution:

Equation of the circle is

x^{2} + y^{2} – 4x + 6y + a = 0

2g = – 4, 2f = 6, c = a

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

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

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

13 – a = 16

a = 13 – 16 = -3

Question 9.

Find the value of ‘a’ if 2x^{2} + ay^{2} – 3x + 2y – 1 =0 represents a circle and also find its radius.

Solution:

General equation of second degree

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

Represents a circle, when

a = b, h = 0, g^{2} + f^{2} – c ≥ 0

In 2x^{2} + ay^{2} – 3x + 2y – 1 = 0

a = 2, above equation represents circle.

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

2g = – \(\frac{3}{2}\); 2f = 1; c = – \(\frac{1}{2}\)

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

Radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}\)

= \(\frac{\sqrt{21}}{4}\) units

Question 10.

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 root of y^{2} + 2py – q^{2} = 0, then find the equation of a circle for which \(\overline{\mathrm{AB}}\) is a diameter. [Mar. 14]

Solution:

Equation of the circle is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – 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_{1}, x_{2} are roots of x^{2} + 2ax – b^{2} = 0

y_{1}, y_{2} are roots of y^{2} + 2py – q^{2} = 0

x_{1} + x_{2} = – 2a

x_{1}x_{2} = -b^{2}

y_{1} + y_{2} = -2p

y_{1}y_{2} = -q^{2}

Equation of circle be

x^{2} – x (- 2a) – b^{2} + y^{2}– y (- 2p) – q^{2} = 0

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

Question 11.

Find the equation of a circle which passes through (4, 1) (6, 5) and having the centre on 4x + 3y – 24 = 0 [A.P. Mar. 16; Mar. 14]

Solution:

Equation of circle be x^{2} + y^{2} + 2gx + 2fy + c = 0 passes through (4, 1) and (6, 5) then

4^{2} + 1^{2} + 2g(4) + 2f(1) + c = 0 ………….. (i)

6^{2} + 5^{2} + 2g(6) + 2f(5) + c = 0 ……………… (ii)

Centre lie on 4x + 3y – 24 = 0

∴ 4(-g) + 3 (-f) – 24 = 0

(ii) – (i) we get

44 + 4g + 8f = 0

Solving (iii) and (iv) we get

f = -4, g = -3, c = 15

∴ Required equation of circle is

x^{2} + y^{2} – 6x – 8y + 15 = 0

Question 12.

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

Solution:

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

C = (- g, – f)

= (3, 2)

Equation of circle concentric with (i) be

(x – 3)^{2} + (y – 2)^{2} = r^{2}

Passes through (-2, 14)

∴ (- 2 – 3)^{2} + (14 – 2)^{2} = r^{2}

169 = r^{2}

Required equation of circle be

(x – 3)^{2} + (y – 2)^{2} = 169

x^{2} + y^{2} – 6x – 4y – 156 = 0

Question 13.

Find the equation of the circle whose centre lies on the X-axis and passing through (- 2, 3) and (4, 5). [A.P. & T.S. Mar. 15]

Solution:

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

(- 2, 3) and (4, 5) passes through (i)

4 + 9 – 4g + 6f + c = 0 ………………………. (ii)

16 + 25 + 8g + 10f + c = 0 …………………. (iii)

(iii) – (ii) we get

28 + 12g + 4f = 0

f + 3g = – 7

Centre lies on X – axis then f = 0

g = -, \(\frac{7}{3}\), f = 0, c = \(\frac{67}{3}\) -, we get by substituting g; f in equation (ii)

Required equation will be 3(x^{2} + y^{2}) – 14x – 67 = 0

Question 14.

Find the equation of circle passing through (1, 2); (3, – 4); (5, – 6) three points.

Solution:

Equation of circle is

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

1 + 4 + 2g + 4f + c = 0 …………………… (i)

9 + 16 + 6g – 8f + c = 0 …………………… (ii)

25 + 36 + 10g – 12f + c = 0 ………………… (iii)

Subtracting (ii) – (i) we get

20 + 4g – 12f = 0

(or) 5 + g – 3f = 0 ……………….. (iv)

Similarly (iii) – (ii) we get

36 + 4g – 4f = 0

(or) 9 + g – f = 0 ………………… (v)

Solving (v) and (iv) we get

f = -2, g = – 11, c = 25

Required equation of circle be x^{2} + y^{2} – 22x – 4y + 25 = 0

Question 15.

Find the length of the chord intercepted by the circle x^{2} + y^{2} – 8x – 2y – 8 = 0 on the line x + y + 1 = 0 [T.S. Mar. 16]

Solution:

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

Centre is C(4, 1), r = \(\sqrt{16+1+8}\) = 5

Equation of the line is x + y + 1 = 0

P = distance from the centre = \(\frac{|4+1+1|}{\sqrt{1+1}}\)

= \(\frac{6}{\sqrt{2}}\) = 3\(\sqrt{2}\)

Length of the chord = 2\(\sqrt{r^{2}-p^{2}}\)

= 2\(\sqrt{25-18}\)

= 2\(\sqrt{7}\) units.

Question 16.

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

Solution:

SS_{11} = S_{1}^{2}

(x^{2} + y^{2} – 2x + 4y – 11) (1 + 9 – 2 + 12 – 11) = [x + 3y – 1 (x + 1) + 2 (y + 3) – 11]^{2}

(x^{2} + y^{2} – 2x + 4y – 11) 9 = [5y – 6]^{2}

9x^{2} + 9y^{2} – 18x + 36y – 99

= 25y^{2} + 36 – 60y

9x^{2} – 16y^{2} – 18x + 96y – 135 = 0

cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\) = \(\frac{|9-16|}{\sqrt{(25)^{2}}}\)

= \(\frac{|-7|}{25}\) = \(\frac{7}{25}\) ⇒ θ = cos^{-1} (\(\frac{7}{25}\))

Question 17.

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. [T.S. Mar. 17]

Solution:

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

Polar of P(4, 2) is

x . 4 + y . 2 – \(\frac{5}{2}\) (x + 4) + 4 (y + 2) + 6 = 0

8x + 4y – 5x – 20 + 8y + 16 + 12 = 0

3x + 12y + 8 = 0

P(4, 2), Q(k, -3) are conjugate points

Polar of P passes through Q

∴ 3k – 36 + 8 = 0

3k = 28 ⇒ k = \(\frac{28}{3}\)

Question 18.

If (2, 0), (0,1), (4, 5) and (0, c) are concyclic, and then find c. [A.P. & T.S. Mar. 15]

Solution:

x^{2} + y^{2} + 2gx + 2fy + c_{1} = 0

Satisfies (2, 0), (0, 1) (4, 5) we get

4 + 0 + 4g + c_{1} = 0 …………………. (i)

0 + 1 + 2g. 0 + 2f + c_{1} = 0 – (ii)

16 + 25 + 8g + 10f + c_{1} = -0 ……………. (iii)

(ii) – (i) we get

– 3 – 4g + 2f = 0

4g – 2f = – 3 ……………… (iv)

(ii) – (iii) we get

– 40 – 8g – 8f = 0 (or)

g + f = – 5 …………………… (v)

Solving(iv) and (v) we get

g = –\(\frac{13}{6}\), f = –\(\frac{17}{6}\)

Substituting g and f values in equation (i) we get

4 + 4 (-\(\frac{13}{6}\)) + c_{1} = 0

c_{1} = \(\frac{14}{3}\)

Now equation x^{2} + y^{2} – \(\frac{13}{3}\) x – \(\frac{17}{3}\) y + \(\frac{14}{3}\) = 0

Now circle passes through (0, c) then

c^{2} – \(\frac{17}{3}\) c + \(\frac{14}{3}\) = 0

3c^{2} – 17c + 14 = 0

⇒ (3c – 14) (c – 1) = 0

(or)

c = 1 or \(\frac{14}{3}\)

Question 19.

Find the length of the chord intercepted by the circle x^{2} + y^{2} – x + 3y – 22 = 0 on the line y = x – 3. [May 11; Mar. 13]

Solution:

Equation of the circle is

S ≡x^{2} + y^{2} – x + 3y – 22 = 0

Centre C(\(\frac{1}{2}\), –\(\frac{3}{2}\))

Question 20.

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

Solution:

Equation of the line is 3x + 4y + 4 = 0

P = Length of the perpendicular .

Length of the chord = 2λ = 2 ⇒ λ = 1

If r is the radius of the circle then

r^{2} = 2^{2} + 1^{2} – 4 + 1 = 5

Centre of the circle is (-2, 3)

Equation of the circle is (x + 2)^{2} + (y – 3)^{2} = 5

x^{2} + 4x + 4 + y^{2} – 6y + 9 – 5 = 0

i.e., x^{2} + y^{2} + 4x – 6y + 8 = 0

Question 21.

Find the pole of 3x + 4y – 45 = 0 with respect to x^{2} + y^{2} – 6x – 8y + 5 = 0. [A.P. Mar. 16]

Solution:

Equation of polar is

xx_{1} + yy_{1} – 3(x + x_{1}) – 4(y + y_{1}) + 5 = 0

x(x_{1} – 3) + y(y_{1} – 4) – 3x_{1} – 4y_{1} + 5 = 0 …………………. (i)

Polar equation is same 3x + 4y – 45 = 0 ……………….. (ii)

Comparing (i) and (ii) we get

Question 22.

i) Show that the circles x^{2} + y^{2} – 6x – 2y + 1 = 0 ; x^{2} + y^{2} + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact. [A.P. Mar. 16; Mar. 11]

Solution:

Equations of the circles are

S_{1} ≡ x^{2} + y^{2} – 6x – 2y + 1 = 0

S_{2} ≡x^{2} + y^{2} + 2x – 8y + 13 = 0

Centres are A (3, 1), B(-1, 4)

r_{1} = \(\sqrt{9+1+1}\) = 3, r_{1} = \(\sqrt{1+16-13}\) = 2

AB = \(\sqrt{(3+1)^{2}+(1-4)^{2}}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5

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

∴ The circles touch each other externally.

The point of contact P divides AB internally in the ratio r_{1} : r_{2} = 3 : 2

Co- ordinates of P are

\(\left(\frac{3(-1)+2.3}{5}, \frac{3.4+2.1}{5}\right)\) i.e., P\(\left(\frac{3}{5}, \frac{14}{5}\right)\)

Equation of the common tangent is S_{1} – S_{2} = 0

-8x + 6y – 12 = 0 (or) 4x – 3y + 6 = 0

ii) Show that x^{2} + y^{2} – 6x – 9y + 13 = 0, x^{2} + y^{2} – 2x – 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.

Solution:

Equations of the circles are

S_{1} ≡ x^{2} + y^{2} – 6x – 9y + 13 = 0

S_{2} ≡ x^{2} + y^{2} – 2x – 16y = 0

centres are A(3, \(\frac{9}{2}\)), B(1, 8)

r_{1} = \(\sqrt{9+\frac{81}{4}-13}\) = \(\frac{\sqrt{65}}{2}\), r_{2} = \(\sqrt{1+64}\)

= \(\sqrt{65}\)

AB = \(\sqrt{(3-1)^{2}+\left(\frac{9}{2}-8\right)^{2}}\) = \(\sqrt{4+\frac{49}{4}}\)

= \(\frac{\sqrt{65}}{2}\)

AB = |r_{1} – r_{2}|

∴ The circles touch each other internally. The point of contact ‘P’ divides AB

externally in the ratio r_{1} : r_{2} = \(\frac{\sqrt{65}}{2}\) : \(\sqrt{65}\)

= 1 : 2 Co-ordinates of P are

\(\left(\frac{1(1)-2(3)}{1-2}, \frac{1(8)-2\left(\frac{9}{2}\right)}{1-2}=\left(\frac{-5}{-1}, \frac{-1}{-1}\right)\right.\) = (5, 1)

p = (5, 1)

∴ Equation of the common tangent is

S_{1} – S_{2} = 0

-4x + 7y + 13 = 0

4x – 7y – 13 = 0

Question 23.

Find the direct common tangents of the circles. [T.S. Mar. 15]

x^{2} + y^{2} + 22x – 4y – 100 = 0 and x^{2} + y^{2} – 22x + 4y + 100 = 0.

Solution:

C_{1} = (-11, 2)

C_{2} = (11, -2)

r_{1} = \(\sqrt{121+4+100}\) = 15

r_{2} = \(\sqrt{121+4-100}\) = 5

Let y = mx + c be tangent

mx – y + c = 0

⊥ from (-11, 2) to tangent = 15

⊥ from (11 ,-2) to tangent = 5

Squaring and cross multiplying

25 (1 + m^{2}) = (11 m + 2 – 22m – 4)^{2}

96m^{2} + 44m – 21 = 0

⇒ 96m^{2} + 72m – 28m – 21 = 0

m = \(\frac{7}{24}\), \(\frac{-3}{4}\)

c = \(\frac{25}{2}\)

y = – \(\frac{3}{4}\)x + \(\frac{25}{2}\)

4y + 3x = 50

c = -22m – 4

= -22(\(\frac{7}{24}\)) – 4

= \(\frac{-77-48}{12}\) =\(\frac{-125}{12}\)

y = \(\frac{7}{24}\) x – \(\frac{125}{12}\)

⇒ 24y = 7x – 250

⇒ 7x – 24y – 250 = 0

Question 24.

Find the transverse common tangents of the circles x^{2} + y^{2} – 4x – 10y + 28 = 0 and x^{2} + y^{2} + 4x-6y + 4 = 0 [A.P. Mar. 15; Mar. 14]

Solution:

C_{1} =(2, 5), C_{2} = (-2, 3)

r_{1} = \(\sqrt{4+25-28}\) = 1,

r_{2} = \(\sqrt{4+9-4}\) = 3

r_{1} + r_{2}= 4

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

= \(\sqrt{16+4}\) = \(\sqrt{20}\)

‘C’ divides C_{1}C_{2} in the ratio 5 : 3

Equation of the pair transverse of the common tangents is

S_{1}^{2} = SS_{11}

(x . 1 + \(\frac{9}{2}\)y – 2(x + 1) – 5(y + \(\frac{9}{2}\)) + 28)^{2} = [1 + \(\frac{81}{4}\) – 4 – 10 × \(\frac{9}{2}\) + 28]

= – (x^{2} + y^{2} – 4x – 10y + 28)

⇒ (-x – \(\frac{1}{2}\)y + \(\frac{7}{2}\))^{2}

= \(\frac{1}{4}\) (x^{2} + y^{2} – 4x – 10y + 28)

(-2x – y + 7)^{2} = (x^{2} + y^{2} – 4x – 10y + 28)

4x^{2} + y^{2} + 4xy – 28x – 14y + 49 = x^{2} + y^{2} – 4x – 10y + 28

3x^{2} + 4xy – 24x – 4y + 21 = 0

(3x + 4y – 21); (x – 1) = 0

3x + 4y – 21 = 0; x – 1 = 0

Question 25.

Show that the circles x^{2} + y^{2} – 4x – 6y – 12 = 0 and x^{2} + y^{2} + 6x + 18y + 26 = 0 touch each other. Also find the point of contact and common tangent at this point of contact. [Mar. 13]

Solution:

Equations of the circles are

x^{2} + y^{2} – 4x – 6y – 12 = 0 and x^{2} + y^{2} + 6x + 18y + 26 = 0

Centres are C_{1}(2, 3), C_{2} = (-3, -9)

r_{1} = \(\sqrt{4+9+12}\) = 5

r_{2} = \(\sqrt{9+81-26}\) = 8

C_{1}C_{2} = \(\sqrt{(2+3)^{2}+(3+9)^{2}}\)

= \(\sqrt{25+144}\) = 13 = r_{1} + r_{2}

∴ Circle touch externally .

Equation of common tangent is S_{1} – S_{2} = 0

-10x -,24y – .38 = 0

5x + 12y + 19 = 0

Question 26.

Find the equation of circle with centre (1, 4) and radius ‘5’.

Solution:

Here (h, k) = (1, 4) and r = 5.

∴ By the equation of the circle with centre at C (h, k) and radius r is

(x – h)^{2} + (y – k)^{2} = r^{2}

(x – 1)^{2} + (y – 4)^{2} = 5^{2}

i.e., x^{2} + y^{2}– 2x – 8y – 8 = 0

Question 27.

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

Solution:

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

g = 1, f = -2, c = -4

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

radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{1+4-(-4)}\) = 3

Question 28.

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

Solution:

Given equation is

3x^{2} + 3y^{2} – 6x + 4y – 4 = 0

Dividing with 3, we have

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

2g = -2, 2f = \(\frac{4}{3}\), c = –\(\frac{4}{3}\)

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

Centre (-g, -f) = (1, \(\frac{-2}{3}\))

radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{1+\frac{4}{9}+\frac{4}{3}}\)

= \(\sqrt{\frac{9+4+12}{9}}\) = \(\sqrt{\frac{25}{9}}\) = \(\frac{5}{3}\)

Question 29.

Find the equation of the circle whose centre is (-1, 2) and which passes through (5, 6).

Solution:

Let C(-1, 2) be the centre of the circle

Since P(5,6) is a point on the circle CP = r

CP^{2} = r^{2} ⇒ r^{2} = (-1 – 5)^{2} + (2 – 6)^{2}

= 36 + 16 = 52

Equation of the circle is (x + 1)^{2} + (y – 2)^{2}

= 52

x^{2} + 2x + 1 + y^{2} – 4y + 4 – 52 = 0

x^{2} + y^{2} + 2x – 4y – 47 = 0

Question 30.

Find the equation of the circle passing through (2, 3) and concentric with the circle x^{2} + y^{2} + 8x + 12y + 15 = 0.

Solution:

The required circle is concentric with the circle x^{2} + y^{2} + 8x + 12y + 15 = 0

∴ The equation of the required circle can be taken as

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

This circle passes through P(2, 3)

∴ 4 + 9 + 16 + 36 + c = 0

c = – 65

Equation of the required circle is x^{2} + y^{2} + 8x + 12y – 65 = 0

Question 31.

From the point A(0, 3) on the circle x^{2} + 4x + (y – 3)^{2} = 0 a chord AB is drawn and extended to a point M such that AM = 2 AB. Find the equation of the locus of M.

Solution:

Let M = (x’, y’)

Given that AM = 2AB

AB + BM = AB + AB

⇒ BM = AB

B is the mid point of AM

Co- ordinates of B are \(\left(\frac{x^{\prime}}{2}-\frac{y^{\prime}+3}{2}\right)\)

B is a point on the circle

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

\(\frac{x^{\prime 2}}{4}\) + 2x’ + \(\frac{y^{\prime 2}-6 y^{\prime}+9}{4}\) = 0

x’^{2} + 8x’ + y’^{2} – 6y’ + 9 = 0

Lotus of M(x’, y’) is x^{2} + y^{2} + 8x – 6y + 9 = 0, which is a circle.

Question 32.

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.

Solution:

Equation of the circle is

x^{2} + y^{2} + ax + by – 12 = 0

Centre = (\(-\frac{a}{2}\), \(-\frac{b}{2}\)) = (2, 3)

\(-\frac{a}{2}\) = 2, \(-\frac{b}{2}\) = 3

a = – 4, b – -6

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

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

Question 33.

If the circle x^{2} + y^{2} – 4x + 6y + a = 0 has radius 4 then find a. [A.P. Mar. 16]

Solution:

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

2g = – 4, 2f = 6, c = a

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

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

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

13 – a = 16

a = 13 – 16 = -3

Question 34.

Find the equation of the circle passing through (4, 1), (6, 5) and having the centre on the line 4x + y – 16 = 0.

Solution:

Let the equation of the required circle be x^{2} + y^{2} + 2gx + 2fy + c = 0 .

This circle passes through A(4, 1)

16 + 1+ 8g + 2f + c = 0

8g + 2f + c = -17 ………………. (1)

The circle passes through B(6, 5)

36 + 25 + 12g + 10f + c = 0

12g + 10f + c = -61 ……………… (2)

The centre (-g, -f) lies on 4x + y – 16 = 0

– 4g – f – 16 = 0

4g + f + 16 = 0 ……………….. (3)

(2) – (1) gives 4g + 8f = -44 ……………….. (4)

4g + f = -16 …………….. (3)

7f = -28

f = \(\frac{-28}{7}\) = -4

From(3) 4g – 4 = -16

4g = -12 ⇒ g = -3

From(1) 8(-3) + 2(-4) + c = -17

c = -17 + 24 + 8 = 15

Equation of the required circle is

x^{2} + y^{2} – 6x – 8y + 15 = 0

Question 35.

Suppose a point (x_{1}, y_{1}) satisfies x^{2} + y^{2}+ 2gx + 2fy + c = 0 then show that it represents a circle whenever g, f and c are real.

Solution:

Comparing with the general equation of second degree co-efficient of x^{2} = coefficient of y^{2} and coefficient of xy = 0

The given equation represents a circle if g^{2} + f^{2} – c ≥ 0

(x_{1}, y_{1}) is a point on the given equation

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

x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c = 0

g^{2} + f^{2} – c = g^{2} + f^{2} + x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} = 0

= (x_{1} + g)^{2} + (y_{1} + f)^{2} ≥ 0

g, f and c are real

∴ The given equation represents a circle.

Question 36.

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

Solution:

Here (x_{1}, y_{1}) = (1, 2) and (x_{2}, y_{2}) = (4, 5)

Equation of the required circle is

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

x^{2} – 5x + 4 + y^{2} – 7y + 10 = 0

x^{2} + y^{2} – 5x – 7y + 14 = 0

Question 37.

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).

Solution:

A = (2, 3) and AB is the diameter of the circle x^{2} + y^{2} – 8x – 8y + 27 = 0

Centre of the circle is C = (4, 4)

Suppose B(x, y) is the other end

C = mid point of AB = \(\left(\frac{2+x}{2}, \frac{3+y}{2}\right)\) = (4, 4)

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

2 + x = 8

x = 6

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

3 + y = 8

y = 5

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

Question 38.

Find the equation of the circum – circle of the traingle formed by the line ax + by + c = 0 (abc ≠ 0) and the co-ordinate axes.

Solution:

Let the line ax + by + c = 0 cut X, Y axes at A and B respectively co-ordinates of O are (0, 0) A are

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

This circle passes through 0(0, 0)

∴ c = 0

This circle passes through A(-\(\frac{c}{a}\), 0)

\(\frac{c^{2}}{a^{2}}\) + 0 – 2\(\frac{\mathrm{gc}}{\mathrm{a}}\) = 0

2g . \(\frac{c}{a}\) = \(\frac{c^{2}}{a^{2}}\) ⇒ 2g = \(\frac{c}{a}\) ⇒ g = \(\frac{c}{2a}\)

The circle passes through B (0, –\(\frac{c}{b}\))

0 + \(\frac{c^{2}}{b^{2}}\) + 0 – 2g \(\frac{c}{b}\) = 0

2f\(\frac{c}{b}\) = \(\frac{c^{2}}{b^{2}}\) ⇒ 2g = \(\frac{c}{b}\) ⇒ f = \(\frac{c}{2b}\)

Equation of the circle, through O, A, B is

x^{2} + y^{2} + \(\frac{c}{a}\) x + \(\frac{c}{b}\) y = 0

ab(x^{2} + y^{2}) + (bx + ay) = 0

This is the equation of the circum circle of ∆OAB

Question 39.

Find the equation of the circle which passes through the vertices of the triangle formed by L_{1} = x + y + 1 = 0; L_{2} = 3x + y- 5 = 0 and L_{3} = 2x + y – 5 = 0.

Solution:

Suppose L_{1}, L_{2},: L_{2}, L_{3} and L_{3}, L_{1} intersect in A, B and C respectively.

Consider a curve whose equation is

k (x + y + 1) (3x + y – 5) + l(3x + y – 5) (2x + y – 5) + m(2x + y – 5) (x + y + 1) = 0 ………………. (1)

This equation represents a circle

i) Co-efficient of x^{2} = Co – efficient of y^{2}

3k + 6l + 2m = k + l + m

2k + 5l + m = 0 ……………….. (2)

ii) Co-efficient of xy = 0

4k + 5l + 3m = 0 ……………….. (3).

Applying cross multiplication rule for (2) and (3) we get

Substituting in (1), equation of the required circle is

5(x + y + 1) (3x + y – 5) – 1 (3x + y – 5)

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

i.e., x^{2} + y^{2} – 30x – 10y + 25 = 0

Question 40.

Find the centre of the circle passing through the points (0, 0), (2, 0)and (0, 2).

Solution:

Here (x_{1}, y_{1}) = (0, 0); (x_{2}, y_{1}) = (2, 0);

(x_{3}, y_{3}) = (0, 2)

c_{1} = -(x_{1}^{2} + y_{1}^{2}) = 0

c_{2} = – (x_{2}^{2} + y_{2}^{2}) = -(2^{2} + 0^{2}) = -4

c_{3} = -(x_{3}^{2} + y_{3}^{2}) = -(0^{2} + 2^{2}) – 4

The centre of the circle passing through three non-collinear points P(x_{1}, y_{1}), Q(x_{2}, y_{2}) and R(x_{3}, y_{3})

Thus the centre of the required circle is (1, 1)

Question 41.

Obtain the parametric equations of the circle x^{2} + y^{2} = 1.

Solution:

Equation of the circle is x^{2} + y^{2} = 1

Centre is (0, 0) radius = r = T

The circle having radius r is x = r cos θ,

y = sin θ where 0 < θ < 2π

The parametric equation of the circle

x^{2} + y^{2} = 1 and

x = 1 . cos θ = cos θ

y = 1 . sin θ = sin θ, θ < θ < 2π

Note: Every point on the circle can be expressed as (cos θ, sin θ)

Question 42.

Obtain the parametric equation of the circle represented by

x^{2} + y^{2} + 6x + 8y – 96 = 0.

Solution:

Centre (h, k) of the circle is (-3, -4)

radius = r = \(\sqrt{9+16+96}\) = \(\sqrt{121}\) = 11

Parametric equations are

x = h + r cos θ = -3 + 11 cos θ

y = k + r sin θ = -4 + 11 sin θ

where 0 < θ < 2π

Question 43.

Locate the position of the point (2, 4) with respect to the circle. x^{2} + y^{2} – 4x – 6y + 11 = 0.

Solution:

Here (x_{1}, y_{1}) = (2, 4) and

S ≡ x^{2} + y^{2} – 4x – 6y + 11

S_{11} = 4 + 16 – 8 – 24 + 11

= 31 – 32 = – 1 < 0

∴ The point (2, 4) lies inside the circle S = 0

Question 44.

Find the length of the tangent from (1, 3) to the circle x^{2} + y^{2} – 2x + 4y – 11 = 0.

Solution:

Here (x_{1}, y_{1}) = (1, 3) and

S = x^{2} + y^{2} – 2x + 4y – 11 = 0

P(x_{1}, y_{1}) to S = 0 is \(\sqrt{S_{11}}\)

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

= \(\sqrt{1+9-2+12-11}\) = \(\sqrt{9}\) = 3

Question 45.

If a point P is moving such that the length of tangents drawn from P to

x^{2} + y^{2} – 2x + 4y – 20 = 0 ……………… (1)

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

are in the ratio 2 : 1.

Then show that the equation of the locus of P is x^{2} + y^{2} – 2x – 12y + 8 = 0.

Solution:

Let P(x_{1}, y_{1}) be any point on the locus and \(\overline{\mathrm{PT}_{1}}\), \(\overline{\mathrm{PT}_{2}}\) be the lengths of tangents from P to the circles (1) and (2) respectively.

x^{2} + y^{2} – 2x + 4y – 20 = 0 and

x^{2} + y^{2} – 2x – 8y + 1 = 0

\(\frac{\overline{\mathrm{PT}_{1}}}{\overline{\mathrm{PT}_{2}}}=\frac{2}{1}\)

i.e., \(\sqrt{x_{1}^{2}+y_{1}^{2}-2 x_{1}+4 y_{1}-20}\)

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

3 (x_{1}^{2} + y_{1}^{2}) – 6x_{1} – 36y_{1} + 24 = 0

Locus of P (x_{1}, y_{1}) is

x^{2} + y^{2} – 2x – 12y + 8 = 0

Question 46.

If S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0. represents a circle then show that the straight line lx + my + n = 0

i) touches the circle S = 0 if

(g^{2} + f^{2} – c) = \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

ii) meets the circle S = 0 in two points if

g^{2} + f^{2} – c > \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

iii) will not meet the circle if

g^{2} + f^{2} – c < \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

Solution:

Let ‘c’ be the centre and ‘r’ be the radius of the circle S = 0

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

i) The given straight line touches the circle

if r = \(\frac{|l(-\mathrm{g})+\mathrm{m}(-\mathrm{f})-\mathrm{n}|}{\sqrt{l^{2}+\mathrm{m}^{2}}}\)

\(\sqrt{g^{2}+f^{2}-c}\) = \(\frac{|-(l g+m f-n)|}{\sqrt{l^{2}+m^{2}}}\)

squaring on both sides, we get

g^{2} + f^{2} – c = \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

ii) The given line lx + my + n = 0 meets the circle s = 0 in two points if

g^{2} + f^{2} – c > \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

iii) The given line lx + my + n = 0 will not meet the circle s = 0 if

g^{2} + f^{2} – c < \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

Question 47.

Find the length of the chord intercepted by the circle x^{2} + y^{2} + 8x – 4y – 16 = 0 on the line 3x – y + 4 = 0.

Solution:

The centre of the given circle c = (-4, 2) and radius r = \(\sqrt{16+4+16}\) = 6. Let the perpendicular distance from the centre to the line 3x-y + 4 = 0 be ‘d’ then

d = \(\frac{|3(-4)-2+4|}{\sqrt{3^{2}+(-1)^{2}}}\) = \(\frac{10}{\sqrt{10}}\) = \(\sqrt{10}\)

Length of the chord + \(\sqrt{r^{2}-d^{2}}\)

= 2\(\sqrt{6^{2}-(\sqrt{10})^{2}}\) = 2\(\sqrt{26}\)

Question 48.

Find the equation of tangents to x^{2} + y^{2} – 4x + 6y – 12 = 0 which are parallel to x + 2y- 8 = 0.

Solution:

Here g = -2, f = 3, r = \(\sqrt{4+9+12}\) = 5

and the slope of the required tangent is \(\frac{-1}{2}\)

The equations of tangents are

y + 3 = \(\frac{-1}{2}\) (x – 2) ± 5 \(\sqrt{1+\frac{1}{4}}\)

2(y + 3) = – x + 2 ± 5\(\sqrt{5}\)

x + 2y + (4 ± 5\(\sqrt{5}\)) = 0

Question 49.

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:

i) We know that the intercept made by S = 0 on X – axis is 2\(\sqrt{g^{2}-c}\)

If the circle touches the X – axis then

22\(\sqrt{g^{2}-c}\) 0 ⇒ g^{2} = c

Question 50.

Find the equation of the tangent to x^{2} + y^{2} – 6x + 4y – 12 = 0 at (- 1, 1).

Solution:

Here (x_{1}, y_{1}) = (-1, 1) and

S ≡ x^{2} + y^{2} – 6x + 4y – 12 = 0

∴ The equation of the tangent is

x(-1) + y (1) – 3(x – 1) + 2(y + 1)- 12 = 0

The equation of the tangent at the point P(_{1}, y_{1}) to the circle

S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0 is S_{1} = 0

⇒ – x + y – 3x + 3 + 2y + 2 – 12 = 0

⇒ – 4x + 3y – 7 = 0

(or) 4x – 3y + 7 = 0

Question 51.

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.

Solution:

Here (x_{1}, y_{1}) = (3, -1) and

S ≡ x^{2} + y^{2} – 2x + 4y = 0

The equation of the tangent at (3, -1) is

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

3x – y – x – 3 + 2y – 2 = 0

2x + y – 5 = 0

Slope of the tangent is m = -2, for the circle

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

r = \(\sqrt{1+4-0}\) = \(\sqrt{5}\)

Equations of the tangents are

y = mx ± r \(\sqrt{1+m^{2}}\) is a tangent to the circle x^{2} + y^{2} = r^{2} and the slope of the tangent is m.

y + 2 = -2(x – 1) ± \(\sqrt{5} \sqrt{1+4}\)

y + 2 = -2x + 2 ± 5

2x 4- y = ± 5

The tangents are

2x + y + 5 = 0 and 2x + y – 5 = 0

The tangent parallel to the given tangent is 2x + y ± 5 = 0

Question 52.

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

Solution:

Let (x_{1}, y_{1}) be the point of contact

Equation of the tangent is

(x_{1} + g) x + (y_{1} + f) y + (gx_{1} + fy_{1} + c) = 0

We have \(\frac{x_{1}-3}{4}\) = \(\frac{y_{1}+2}{-3}\)

= \(\left(\frac{-3 x_{1}+2 y_{1}-12}{7}\right)\) ……………… (1)

From first and second equalities of (1), we get

3x_{1} + 4y_{1} = 1 ………………. (2)

Now by taking first and third equalities of (1), we get

19x_{1} – 8y_{1} = -27 ………………. (3)

Solving (2) and (3) we obtain

x_{1} = -1;, y_{1} = 1

Hence the point of contact is (-1, 1).

Question 53.

Find the equations of circles which touch 2x – 3y + 1 = 0 at (1, 1) and having radius \(\sqrt{13}\).

Solution:

The centres of the required circle lies on a line perpendicular to 2x – 3y + 1 =0 and passing through (1, 1)

The equation of the line of centre can be taken as

3x + 2y + k = 0

This line passes through (1, 1)

3 + 2 + k = 0 ⇒ k = -5

Equation of AB is 3x + 2y – 5 = 0

The centres A and B are situated on

3x + 2y – 5 = 0 at a distance \(\sqrt{13}\) from (1, 1).

The centres are given by

(x_{1} ± r cos θ, y_{1} ± r sin θ)

\(\left(1+\sqrt{13}\left(-\frac{2}{\sqrt{13}}\right) 1+\sqrt{13} \cdot \frac{3}{\sqrt{13}}\right)\) and

\(\left(1-\sqrt{13} \frac{(-2)}{\sqrt{13}}, 1-\sqrt{13} \cdot \frac{3}{\sqrt{13}}\right)\)

i.e., (1 -2, 1 +3) and (1 + 2, 1 – 3)

(-1, 4) and (3, -2)

Centre (3, -2), r = \(\sqrt{13}\)

Equation of the required circles are

(x + 1 )^{2} + (y – 4)^{2} = 13 and

(x – 3)^{2} + (y + 2)^{2} = 13

i.e., x^{2} + y^{2} + 2x – 8y + 4 = 0

and x^{2} + y^{2} – 6x + 4y = 0

Question 54.

Show that the line 5x + 12y – 4 = 0 touches the circle x^{2} + y^{2} – 6x + 4y + 12 = 0.

Solution:

Equation of the circle is

x^{2} + y^{2} – 6x + 4y + 12 = 0

Centre (3, -2), r = \(\sqrt{9+4-12}\) = 1 ……………….. (1)

The given line touches the circle if the perpendicular distance from the centre on the given line is equal to radius of the circle, d = perpendicular distance from (3, -2)

= \(\frac{|5(3)+12(-2)-4|}{\sqrt{25+144}}\)

= \(\frac{13}{13}\) = 1 = radius of the circle ………………… (2)

∴ The given line 5x + 12y – 4 = 0 touches the circle.

Question 55.

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:

Here g = -3, f = -2; r = \(\sqrt{9+4-12}\) = 5

∴ The equation of the chord joining the points θ_{1} = 30°, θ_{2} = 60°

Equation of chord joining the point; (-g+ r cos θ_{1}(-f + r sin θ_{1}) where r is the radius of the circle; θ_{2} and (-g + r cos θ_{2}, -f + r sin θ_{2}) is (x + g) cos (\(\frac{\theta_{1}+\theta_{2}}{2}\)) + (y + f)

sin (\(\frac{\theta_{1}+\theta_{2}}{2}\)) = r cos (\(\frac{\theta_{1}+\theta_{2}}{2}\))

(\(\frac{\theta_{1}+\theta_{2}}{2}\)) = r cos \(\frac{\theta_{1}+\theta_{2}}{2}\)

(x – 3) cos [latex]\frac{60^{\circ}+30^{\circ}}{2}[/latex]

(y + 2) sin [latex]\frac{60^{\circ}+30^{\circ}}{2}[/latex]

= 5 cos [latex]\frac{60^{\circ}-30^{\circ}}{2}[/latex]

i.e., (x – 3) cos 45 ° + (y + 2) sin 45°

= 5 cos 15°

= \(\frac{(x-3)+(y+2)}{\sqrt{2}}=5\left[\frac{(\sqrt{3}+1)}{2 \sqrt{2}}\right]\)

i.e., 2x + 2y – (7 + 5\(\sqrt{3}\)) = 0.

Question 56.

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

Solution:

Equation of the circle is

x^{2} + y^{2} + 4x + 6y – 39 = 0

g = 2,f = 3, r = \(\sqrt{4+9+39}\) = \(\sqrt{52}\) = 2\(\sqrt{3}\)

θ = 30°

Equation of the tangent is

(x + g) cos 30° + (y + f) sin 30° = r

(x + 2) \(\frac{\sqrt{3}}{2}\) +(y + 3) \(\frac{1}{2}\) = 2713

\(\sqrt{3}\)x + 2\(\sqrt{3}\) + y + 3 = 4\(\sqrt{13}\)

\(\sqrt{3}\) x + y + (3 + 2\(\sqrt{3}\) – 4\(\sqrt{13}\)) = 0

Question 57.

Find the area of the triangle formed by the tangent at P(x_{1}, y_{1}) to the circle x^{2} + y^{2} = a^{2} with the co-ordinate axes where x_{1}y_{1} ≠ 0.

Solution:

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

Equation of the tangent at P(x_{1}, y_{1}) is xx_{1} + yy_{1} = a^{2} ……………… (1)

= \(\frac{a^{4}}{2\left|x_{1} y_{1}\right|}\)

Question 58.

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:

Let A(3, 2), C be the centre of given circle and the normal at A meet the circle at B(a, b). From the hypothesis, we have

2g = -4 i.e., g = -2

2f = -6 i.e., f = -3

x_{1} = 3 and y_{1} = 2

The equation of the normal at P(x_{1}, y_{1}) of the circle

S ≡ x^{2} + y^{2} + 2gx + 2fy + c = Q is

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

The equation of normal at A(3, 2) is

(x – 3) (2 – 3) – (y – 2) (3 – 2) = 0

i.e., x + y – 5 = 0

The centre of the circle is the mid point of A and B. (Points of intersection of normal and circle).

\(\frac{a+3}{2}\) = 2 ⇒ a = 1

and \(\frac{b+2}{2}\) = 3 ⇒ b = 4

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

Question 59.

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 co-ordinate axes.

Solution:

Here 2g = -22, 2f = – 4, g = -11, f = -2

x_{1} = 3, y_{1} = – 4

Equation of the normal at (3, -4) is

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

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

3x + 4y – 25 = 0 ……………….. (1)

This line meets X-axis at A(\(\frac{25}{3}\), 0) and Y – axis at B(0, \(\frac{25}{4}\)) , ∆OAB = \(\frac{1}{2}\) |OA . OB|

= \(\frac{1}{2}\) |\(\frac{25}{3} \times-\left[\frac{25}{4}\right]\)| = \(\frac{625}{24}\)

Question 60.

Show that the line lx + my + n =0 is a normal to the circle S = 0 if and only if gl + 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 (-g, -f) lies on

lx + my + n = 0

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

gl + fm = n

Question 61.

Find the condition that the tangents drawn from the exterior point (g, f) to S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0 are perpendicular to each other.

Solution:

If the angle between the tangents drawn from P(x_{1}, y_{1}) to S = 0 is θ, then

tan (\(\frac{\theta}{2}\)) = \(\frac{\mathrm{r}}{\sqrt{\mathrm{S}_{11}}}\)

Equation of the circle is

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

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

S_{11} = g^{2} + f^{2} + 2g^{2} + 2f^{2} + c

= 3g^{2} + 3f^{2} + c

θ = 90° ⇒ tan \(\frac{\theta}{2}\) = tan 45 = \(\frac{\sqrt{g^{2}+f^{2}-c}}{\sqrt{3 g^{2}+3 f^{2}+c}}\)

1 = \(\frac{g^{2}+f^{2}-c}{3 g^{2}+3 f^{2}+c}\)

⇒ 3g^{2} + 3f^{2} + c = g^{2} + f^{2} – c

⇒ 2g^{2} + 2f^{2} + 2c = 0 ⇒ g^{2} + f^{2} + c = 0

This is the condition for the tangent drawn from (g, f) to the circle S = 0 are perpendicular.

Note : Here c < 0

Question 62.

If θ_{1}, θ_{2} are the angles of inclination of tangents through a point P to the circle x^{2} + y^{2} = a^{2} then find the locus of P when cot θ_{1} + cot θ_{2} = k.

Solution:

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

If m is the slope of the tangent, then the equation of tangent passing through P(x_{1}, y_{1}) can be taken as

y_{1} = mx_{1} ± a\(\sqrt{1+m^{2}}\)

(y_{1} – mx_{1})^{2} = a^{2} (1 + m^{2})

m^{2}x_{1}^{2} + y_{1}^{2} – 2mx_{1}y_{1} – a^{2} – a^{2}m^{2} = 0

m^{2} (x_{1}^{2} – a^{2}) – 2mx_{1}y_{1} + (y_{1}^{2} – a^{2}) = 0

Suppose m_{1} and m_{2} are the roots of this equation

m_{1} + m_{2} = tan θ_{1} + tan θ_{2} = \(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}\)

m_{1}m_{2} – tan θ_{1} . tan θ_{1} = \(\frac{y_{1}^{2}-a^{2}}{x_{1}^{2}-a^{2}}\)

Given that cot θ_{1} + cot θ_{2} = k

⇒ \(\frac{1}{\tan \theta_{1}}+\frac{1}{\tan \theta_{2}}\) = k

⇒ \(\frac{\tan \theta_{1}+\tan \theta_{2}}{\tan \theta_{1} \cdot \tan \theta_{2}}\) = k

⇒ \(\frac{2 x_{1} y_{1}}{y_{1}^{2}-a^{2}}\) = k

2x_{1}y_{1} = k (y_{1}^{2} – a^{2})

Locus of P(x_{1}, y_{1}) is 2xy = k(y^{2} – a^{2})

Also, conversely if P(x_{1}, y_{1}) satisfies the condition 2xy = k(y^{2} – a^{2}) then it can be show that cot θ_{1} + cot θ_{2} = k

Thus the locus of P is 2xy = k(y^{2} – a^{2})

Question 63.

Find the chord of contact of (2, 5) with respect to the circle x^{2} + y^{2} – 5x + 4y – 2 = 0.

Solution:

Here (x_{1}, y_{1}) = (2, 5). By

S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0 then the equation of the chord of contact of P with respect to S = 0 is S_{1} =0, the required chord of contact is

xx_{1} + yy_{1} – \(\frac{5}{2}\) (x + x_{1}) + 2(y + y_{1}) – 2 = 0

Substituting x_{1} and y_{1} values, we get

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

i.e., x – 14y + 6 = 0.

Question 64.

If the chord of contact of a point P with respect to the circle x^{2} + y^{2} = a^{2} cut the circle at A and B such that, AÔB = 90° then show that P lies on the circle x^{2} + y^{2} = 2a^{2}.

Solution:

Given circle x^{2} + y^{2} = a^{2} …………… (1)

Let P(x_{1}, y_{1}) be a point and let the chord of contact of it cut the circle in A and B such that AÔB = 90°. The equation of the chord of contact of P(x_{1}, y_{1}) with respect to (1) is

xx_{1} + yy_{1} – a^{2} = 0 ……………………. (2)

The equation of the pair of the lines \(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) is.given by x^{2} + y^{2} – a^{2}

\(\left(\frac{x x_{1}+y y_{1}}{a^{2}}\right)^{2}\) = 0

or a^{2} (x^{2} + y^{2})- (xx_{1} + yy_{1})^{2} = 0

or x^{2} (a^{2} – x_{1}^{2}) – 2x_{1}y_{1}xy + y^{2} (a^{2} – y_{1}^{2}) = 0 – (3)

Since AÔB = 90°, we have the coefficient of x^{2} in (3) + coefficient of y^{2} in (3) = 0

∴ a^{2} – x_{1}^{2} + a^{2} – y_{1}^{2} = 0

i.e., x_{1}^{2} + y_{1}^{2} = 2a^{2}

Hence the point P(x_{1}, y_{1}) lies on the circle x^{2} + y^{2} = 2a^{2}.

Question 65.

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

Solution:

Here (x_{1}, y_{1}) = (2, 3) ⇒ x_{1} = 2, y_{1} = 3

Equation of the circle is

x^{2} + y^{2} + 6x + 8y – 96 = 0

Equation of the polar is S_{1} = 0

Polar of (2, 3) is 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 66.

Find the pole of x + y + 2 = 0 with respect to the circle x^{2} + y^{2} – 4x + 6y – 12 = 0.

Solution:

Here lx + my + n = 0 is x + y + 2 = 0

l = 1, m = 1, n = 2

Equation of the circle is

S ≡ x^{2} + y^{2} – 4x + 6y – 12 = 0

2g = -4, 2f = 6, c = -12

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

The pole of the given line x + y + 2 = 0 w.r.to the given circle (-23, -28)

Question 67.

Show that the poles of the tangents to the circle x^{2} + y^{2} = a^{2} with respect to the circle (x + a)^{2} + y^{2} = 2a^{2} lie on y^{2} + 4ax = 0.

Solution:

Equations of the given circles are

x^{2} + y^{2} = a^{2} …………………. (1)

and (x + a)^{2} + y^{2} = 2a^{2} ………………. (2)

Let P(x_{1}, y_{1}) be the pole of the tangent to the circle (1) with respect to circle (2).

The polar of P w.r.to circle (2) is

xx_{1} + yy_{1} + a(x + x_{1}) – a^{2} = 0

x(x_{1} + a) + yy_{1} + (ax_{1} – a_{2}) = 0

This is a tangent to circle (1)

∴ a = \(\frac{\left|0 .+0+a x_{1}-a^{2}\right|}{\sqrt{\left(\dot{x}_{1}+a\right)^{2}}+y_{1}^{2}}\)

a = \(\frac{a\left|x_{1}-a\right|}{\sqrt{\left(x_{1}+a\right)^{2}}+y_{1}^{2}}\)

Squaring and cross – multiplying

(x_{1} + a)^{2} + y_{1}^{2} = (x_{1} – a)^{2}

(or) y_{1}^{2} + (x_{1} + a)^{2} – (x_{1} – a)^{2} = 0

i.e., y_{1}^{2} + 4ax_{1} = 0

The poles of the tangents to circle (1) w.r.to circle (2) lie on the curve y^{2} + 4ax = 0

Question 68.

Show that (4, -2) and (3, -6) are conjugate with respect to the circle x^{2} + y^{2} – 24 = 0.

Solution:

Here (x_{1}, y_{1}) = (4, -2) and (x_{2}, y_{2}) = (3, -6) and

S ≡ x^{2} + y^{2} – 24 = 0

Two points (x_{1}, y_{1}) and (x_{2}, y_{2}) are conjugate with respect to the circle S = 0 if S_{12} = 0;

In this case x_{1}x_{2} + y_{1}y_{2} – 24 = 0

S_{12} = 4.3 + (-2) (-6) – 24 .

= 12 + 12 – 24 = 0

∴ The given points are conjugate with respect to the given circle.

Question 69.

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

Solution:

Here (x_{1}, y_{1}) = (4, k) and (x_{2}, y_{2}) = (2, 3) and

S ≡ x^{2} + y^{2} – 17 = 0

The given points are conjugate ⇒ S_{12} = 0

x_{1}x_{2} + y_{1}y_{2} – 17 = 0

4.2 + k. 3 – 17 = 0

3k = 17 – 8 = 9

k = \(\frac{9}{3}\) = 3

Question 70.

Show that the lines 2x + 3y + 11 = 0 and 2x – 2y – 1 = 0 are conjugate with respect to the circle x^{2} + y^{2} + 4x + 6y – 12 = 0.

Solution:

Here l_{1} = 2, m_{1} = 3, n_{1} = 11

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

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

r = \(\sqrt{9+4-12}\) = 1

We know that l_{1}x + m_{1}y + n_{1} = 0

l_{2}x + m_{2}y + n_{2} = 0 are conjugate with respect to S = 0

r^{2} (l_{1}l_{2} + m_{1}m_{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}) – 1(2.2 + 3(- 2)) = 4 – 6 = -2

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

= (2.2 + 3.3-11) (2.2-2.3 + 11)

= 2(- 1) = -2 L.H.S. = R.H.S.

Condition for conjugate lines is satisfied

∴ Given lines are conjugate lines.

Question 71.

Show that the area of the triangle formed by the two tangents through P(xv 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 \(\frac{r\left(S_{11}\right)^{3 / 2}}{S_{11}+r^{2}}\) where r is the radius of the circle.

Solution:

PA and PB are two tangents from P to the circle S = 0

AB is the chord of contact

= \(S_{11} \cdot \frac{r}{\sqrt{S_{11}}} \cdot \frac{S_{11}}{S_{11}+r^{2}}\)

= \(\frac{r\left(S_{11}\right)^{3 / 2}}{S_{11}+r^{2}}\)

Question 72.

Find the mid point of the chord intercepted by

x^{2} + y^{2} – 2x – 10y + 1 = 0 ………………. (1)

on the line x – 2y + 7 = 0. ……………. (2)

Solution:

Let P(x_{1}, y_{1}) be the mid point of the chord AB

Equation of the chord is S_{1} = S_{11}

xx_{1} + yy_{1} – 1 (x + x_{1}) – 5(y + y_{1}) + 1 = x_{1}^{2} + y_{1}^{2} – 2x_{1} – 10y_{1} + 1 .

x(x_{1} – 1) + y(y_{1} – 5) – (x_{1}^{2} + y_{1}^{2} – x_{1} – 5y_{1}) = 0 ………………. (3)

Equation of the given line is x – 2y – 7 = 0

Comparing (1) and (2)

\(\frac{x_{1}-1}{1}=\frac{y_{1}-5}{-2}=\frac{x_{1}^{2}+y_{1}^{2}-x_{1}-5 y_{1}}{-7}\) = k (say)

x_{1} – 1 = k ⇒ x_{1} = k + 1

y_{1} – 5 = -2(k) ⇒ y_{1} = 5 – 2k

x_{1}^{2} + y_{1}^{2} – x_{1} – 5y_{1} = -7k

⇒ (k + 1)^{2} + (5 – 2k)^{2} – (k + 1) – 5(5 – 2k) + 7k = 0

⇒ k^{2} + 2k + 1 + 25 + 4k^{2} – 20k – k – 1 – 25 + 10k + 7k = 0

⇒ 5k^{2} – 2k = 0

⇒ k (5k — 2) = 0 ⇒ k = 0, \(\frac{2}{5}\)

k = 0 ⇒ (x_{1}, y_{1}) = (1, 5) and x – 2y + 7

= 1 – 10 + 7 ≠ 0

(1, 5) is not a point on the chord.

k = \(\frac{2}{5}\) (x_{1}, y_{1}) = \(\left(\frac{7}{5}, \frac{21}{5}\right)\)

Question 73.

Find the locus of mid-points of the chords of contact of x^{2} + y^{2} = a^{2} from the points lying on the line lx + my + n = 0.

Solution:

Let P = (x_{1}, y_{1}) be a point on the locus P is the mid point of the chord of the circle

x^{2} + y^{2} = a^{2}

Equation of the chord is lx + my + n = 0 ……………… (1)

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

Equation is the chord having (x_{1}, y_{1}) as mid point of S_{1} = S_{11}

xx_{1} + yy_{1} = x_{1}^{2} + y_{1}^{2}

xx_{1} + yy_{1} – (x_{1}^{2} + y_{1}^{2}) = 0 ……………….. (2)

Pole of (2) with respect to the circle in

Locus of P(x_{1}, y_{1}) is n(x^{2} + y^{2}) + a^{2}(lx + my) = 0

Question 74.

Show that the four common tangents can be drawn for the circles given by

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

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

and find the internal and external centres of similitude. [T.S. Mar. 19]

Solution:

Equations of the circles are

x^{2} + y^{2} – 14x + 6y + 33 = 0

and x^{2} + y^{2} + 30x – 2y + 1 = 0

Centres are A(7, -3), B(-15, 1)

r_{1} = \(\sqrt{49+9-33}\) = 5

r_{2} = \(\sqrt{225+1-1}\) = 15

AB = \(\sqrt{(7+15)^{2}+(-3-1)^{2}}\)

= \(\sqrt{484+16}\) = \(\sqrt{500}\) > r_{1} + r_{2}

∴ Four common tangents can be drawn to the given circle

r_{1} : r_{2} = 5 : 15 = 1 : 3

Internal centre of similitude S’ divides AB in the ratio 1 : 3 internally

Co-ordinates of S’ are

\(\left(\frac{1 .(-15)+3.7}{1+3}, \frac{1.1+3(-3)}{1+3}\right)\)

= \(\left(\frac{6}{4}, \frac{1-9}{4}\right)\) = \(\left(\frac{3}{2},-2\right)\)

External centre of similitude S divides AB externally in the ratio 1 : 3

Co-ordinates of S are

\(\left(\frac{1(-15)+3(7)}{1-3}, \frac{1.1-3(-3)}{1-3}\right)\)

= \(\left(\frac{-15-21}{-2}, \frac{1+9}{-2}\right)\) = (18, -5)

Question 75.

Prove that the circles x^{2} + y^{2} – 8x – 6y + 21 = 0 and x^{2} + y^{2} – 2y – 15 = 0 have exactly two common tangents. Also find the point of intersection of those tangents.

Solution:

Let C_{1}, C_{2} be the centres and r_{1}, r_{2} be their radii.

Equation of the circles are

x^{2} + y^{2} – 8x – 6y + 21 = 0

and x^{2} + y^{2} – 2y – 15 = 0

C_{1}(4, 3), C_{2}(0, 1)

r_{1} = \(\sqrt{16+9-21}\) = 2, r_{2} = \(\sqrt{1+15}\) = 4

\(\overline{\mathrm{C}_{1} \mathrm{C}_{2}^{2}}\) = (4 – 0)^{2} + (3 – 1)^{2} = 16 + 4 = 20

C_{1}C_{2} = 2\(\sqrt{5}\)

|r_{1} – r_{2}| = |2 – 4| = 2, r_{1} + r_{2} = 2 + 4 = 6

|r_{1} – r_{2}| < C_{1}C_{2} < r_{1} + r_{2}

Given circles intersect and have exactly two common tangents.

r_{1} : r_{2} = 2 : 4 = 1 : 2

The tangents intersect in external centre of similitude

Co-ordinates of S are

\(\left(\frac{1.0-2.4}{1-2}, \frac{1.0-2.3}{1-2}\right)\) = \(\left(\frac{-8}{-1}, \frac{-5}{-1}\right)\)

= (8, 5)

Question 76.

Show that the circles x^{2} + y^{2} – 4x – 6y – 12 = 0 and x^{2} + y^{2} + 6x + 18y + 26 = 0 touch each other. Also find the point of contact and common tangent at this point of contact. [Mar. 13]

Solution:

Equations of the circles are

x^{2} + y^{2} – 4x – 6y – 12 = 0 and x^{2} + y^{2} + 6x + 18y + 26 = 0

Centres are C_{1}(2, 3), C_{2} = (-3, -9)

r_{1} = \(\sqrt{4+9+12}\) = 5

r_{2} = \(\sqrt{9+81-26}\) = 8

C_{1}C_{2} = \(\sqrt{(2+3)^{2}+(3+9)^{2}}\)

= \(\sqrt{25+144}\) = 13 = r_{1} + r_{2}

∴ Circle touch externally

Equation of common tangent is S_{1} – S_{2} = 0

-10x – 24y – 38 = 0

5x + 12y + 19 = 0

Question 77.

Show that the circles x^{2} + y^{2} – 4x – 6y – 12 = 0 and 5(x^{2} + y^{2}) – 8x – 14y – 32 = 0 touch each other and find their point of contact.

Solution:

Equations of the circles are

x^{2} + y^{2} – 4x – 6y – 12 = 0

and x^{2} + y^{2} – \(\frac{8}{5}\) x – \(\frac{14}{5}\) y – \(\frac{32}{4}\) = 0

Centres are A(2, 3), B(\(\frac{4}{5}\), \(\frac{7}{5}\))

The circles touch internally

P divides AB externally the ratio 5 : 3

Question 78.

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

Solution:

(x_{1}, y_{1}) = (1o, 4)

Equation of the circle is x^{2} + y^{2} – 25 = 0

Equation of the pair of tangents is S_{1}^{2} = S . S_{11}

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

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

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

Question 79.

Find the equation to all possible common tangents of the circles x^{2} + y^{2} – 2x – 6y + 6 = 0 and x^{2} + y^{2} = 1.

Solution:

Equations of the circles are

x^{2} + y^{2} – 2x – 6y + 6 = 0

and x^{2} + y^{2} = 1

Centres are A(1, 3), B(0, 0),

r_{1} = \(\sqrt{1+9-6}\) = 2

r_{2} = 1

External centre of similitude S divides AB externally in the ratio 2 : 1

Co-ordinates of S are

\(\left(\frac{2.0-1.1}{2-1}, \frac{2.0-1.3}{2-1}\right)\) = (-1, -3)

Equation to the pair of direct common tangents are

(x^{2} + y^{2} – 1) (1 + 9 – 1) = (x + 3y + 1)^{2}

This can be expressed as

(y – 1) (4y + 3x – 5) = 0

Equations of direct common tangents are

y – 1 = 0 and 3x + 4y – 5 = 0

Internal centre of S’ divides AB internally in the ratio 2 : 1

Co-ordinates of S’ are

\(\left(\frac{2.0+1.1}{2+1}=\frac{2.0+1.3}{2+1}\right)\left(\frac{1}{3}, 1\right)\)

Equation to the pair of transverse common tangents are

(\(\frac{x}{3}\) +y – 1)^{2}

= (\(\frac{1}{9}\) + 1 – 1) (x^{2} + y^{2} – 1)

This can be expressed as

(x + 1)(4x – 3y – 5) = 0

Equation of transverse common tangents are x + 1 = 0 and 4x – 3y – 5 = 0