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 x2 + y2 + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3), find g, f and its radius. [Mar. 11]
Solution:
Circle is x2 + y2 + 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 x2 + y2 = 4 [Mar. 14]
Solution:
Equation of the circle is x2 + y2 = 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 = 82 [A.P. Mar. 16; Mar. 11]
Solution:
Equation of the circle is (x – 3)2 + (y – 4)2 =82
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 ≡ x2 + y2 – 6x + 4y – 12
Solution:
Power of the point = S11
= 1 + 1 + 6 + 4 – 12 = 0

iii) P = (2, 3) and S ≡ x2 + y2 – 2x + 8y – 23
Power of the point = S11
= 4 + 9 – 4 + 24 – 23 = 10

iv) P = (2, 4) and S ≡ x2 + y2 – 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 x2 + y2 + 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 x2 + y2 – 10x – 10y + 25 = 0 [T.S. Mar. 15]
Solution:
Equation of the circle is x2 + y2 – 10x – 10y + 25 = 0
Equation of the polar is S1 = 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 x2 + y2 = 35. [T.S. Mar. 16]
Solution:
Equation of the circle is
x2 + y2 = 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 x2 + y2 – 4x + 6y + a = 0 has radius 4 then find a. [A.P. Mar. 16]
Solution:
Equation of the circle is
x2 + y2 – 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 2x2 + ay2 – 3x + 2y – 1 =0 represents a circle and also find its radius.
Solution:
General equation of second degree
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
Represents a circle, when
a = b, h = 0, g2 + f2 – c ≥ 0
In 2x2 + ay2 – 3x + 2y – 1 = 0
a = 2, above equation represents circle.
x2 + y2 – $$\frac{3}{2}$$ 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, x2 + 2ax – b2 = 0 and ordinates of A, B are root of y2 + 2py – q2 = 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 – x1) (x – x2) + (y – y1) (y – y2) = 0
x2 – x(x1 + x2) + x1x2 + y2 – y (y1 + y2) + y1y2 = 0
x1, x2 are roots of x2 + 2ax – b2 = 0
y1, y2 are roots of y2 + 2py – q2 = 0
x1 + x2 = – 2a
x1x2 = -b2

y1 + y2 = -2p
y1y2 = -q2
Equation of circle be
x2 – x (- 2a) – b2 + y2– y (- 2p) – q2 = 0
x2 + 2xa + y2 + 2py – b2 – q2 = 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 x2 + y2 + 2gx + 2fy + c = 0 passes through (4, 1) and (6, 5) then
42 + 12 + 2g(4) + 2f(1) + c = 0 ………….. (i)
62 + 52 + 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
x2 + y2 – 6x – 8y + 15 = 0

Question 12.
Find the equation of a circle which is concentric with x2 + y2 – 6x – 4y – 12 = 0 and passing through (- 2, 14). [Mar. 14]
Solution:
x2 + y2 – 6x – 4y – 12 = 0 …………… (i)
C = (- g, – f)
= (3, 2)
Equation of circle concentric with (i) be
(x – 3)2 + (y – 2)2 = r2
Passes through (-2, 14)
∴ (- 2 – 3)2 + (14 – 2)2 = r2
169 = r2
Required equation of circle be
(x – 3)2 + (y – 2)2 = 169
x2 + y2 – 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:
x2 + y2 + 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(x2 + y2) – 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
x2 + y2 + 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 x2 + y2 – 22x – 4y + 25 = 0

Question 15.
Find the length of the chord intercepted by the circle x2 + y2 – 8x – 2y – 8 = 0 on the line x + y + 1 = 0 [T.S. Mar. 16]
Solution:
Equation of the circle is x2 + y2 – 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 x2 + y2 – 2x + 4y – 11 =0 and also find the angle between them. [T.S. Mar. 16]
Solution:
SS11 = S12
(x2 + y2 – 2x + 4y – 11) (1 + 9 – 2 + 12 – 11) = [x + 3y – 1 (x + 1) + 2 (y + 3) – 11]2
(x2 + y2 – 2x + 4y – 11) 9 = [5y – 6]2
9x2 + 9y2 – 18x + 36y – 99
= 25y2 + 36 – 60y
9x2 – 16y2 – 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 x2 + y2 – 5x + 8y + 6 = 0. [T.S. Mar. 17]
Solution:
Equation of the circle is x2 + y2 – 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:
x2 + y2 + 2gx + 2fy + c1 = 0
Satisfies (2, 0), (0, 1) (4, 5) we get
4 + 0 + 4g + c1 = 0 …………………. (i)
0 + 1 + 2g. 0 + 2f + c1 = 0 – (ii)
16 + 25 + 8g + 10f + c1 = -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 (-$$\frac{13}{6}$$) + c1 = 0
c1 = $$\frac{14}{3}$$
Now equation x2 + y2 – $$\frac{13}{3}$$ x – $$\frac{17}{3}$$ y + $$\frac{14}{3}$$ = 0
Now circle passes through (0, c) then
c2 – $$\frac{17}{3}$$ c + $$\frac{14}{3}$$ = 0
3c2 – 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 x2 + y2 – x + 3y – 22 = 0 on the line y = x – 3. [May 11; Mar. 13]
Solution:
Equation of the circle is
S ≡x2 + y2 – 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
r2 = 22 + 12 – 4 + 1 = 5
Centre of the circle is (-2, 3)
Equation of the circle is (x + 2)2 + (y – 3)2 = 5
x2 + 4x + 4 + y2 – 6y + 9 – 5 = 0
i.e., x2 + y2 + 4x – 6y + 8 = 0

Question 21.
Find the pole of 3x + 4y – 45 = 0 with respect to x2 + y2 – 6x – 8y + 5 = 0. [A.P. Mar. 16]
Solution:
Equation of polar is
xx1 + yy1 – 3(x + x1) – 4(y + y1) + 5 = 0
x(x1 – 3) + y(y1 – 4) – 3x1 – 4y1 + 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 x2 + y2 – 6x – 2y + 1 = 0 ; x2 + y2 + 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
S1 ≡ x2 + y2 – 6x – 2y + 1 = 0
S2 ≡x2 + y2 + 2x – 8y + 13 = 0
Centres are A (3, 1), B(-1, 4)
r1 = $$\sqrt{9+1+1}$$ = 3, r1 = $$\sqrt{1+16-13}$$ = 2
AB = $$\sqrt{(3+1)^{2}+(1-4)^{2}}$$ = $$\sqrt{16+9}$$ = $$\sqrt{25}$$ = 5
AB = 5 = 3 + 2 = r1 + r1
∴ The circles touch each other externally.
The point of contact P divides AB internally in the ratio r1 : r2 = 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 S1 – S2 = 0
-8x + 6y – 12 = 0 (or) 4x – 3y + 6 = 0

ii) Show that x2 + y2 – 6x – 9y + 13 = 0, x2 + y2 – 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
S1 ≡ x2 + y2 – 6x – 9y + 13 = 0
S2 ≡ x2 + y2 – 2x – 16y = 0
centres are A(3, $$\frac{9}{2}$$), B(1, 8)
r1 = $$\sqrt{9+\frac{81}{4}-13}$$ = $$\frac{\sqrt{65}}{2}$$, r2 = $$\sqrt{1+64}$$
= $$\sqrt{65}$$
AB = $$\sqrt{(3-1)^{2}+\left(\frac{9}{2}-8\right)^{2}}$$ = $$\sqrt{4+\frac{49}{4}}$$
= $$\frac{\sqrt{65}}{2}$$
AB = |r1 – r2|
∴ The circles touch each other internally. The point of contact ‘P’ divides AB
externally in the ratio r1 : r2 = $$\frac{\sqrt{65}}{2}$$ : $$\sqrt{65}$$
= 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
S1 – S2 = 0
-4x + 7y + 13 = 0
4x – 7y – 13 = 0

Question 23.
Find the direct common tangents of the circles. [T.S. Mar. 15]
x2 + y2 + 22x – 4y – 100 = 0 and x2 + y2 – 22x + 4y + 100 = 0.
Solution:
C1 = (-11, 2)
C2 = (11, -2)
r1 = $$\sqrt{121+4+100}$$ = 15
r2 = $$\sqrt{121+4-100}$$ = 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 + m2) = (11 m + 2 – 22m – 4)2
96m2 + 44m – 21 = 0
⇒ 96m2 + 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 x2 + y2 – 4x – 10y + 28 = 0 and x2 + y2 + 4x-6y + 4 = 0 [A.P. Mar. 15; Mar. 14]
Solution:
C1 =(2, 5), C2 = (-2, 3)
r1 = $$\sqrt{4+25-28}$$ = 1,
r2 = $$\sqrt{4+9-4}$$ = 3
r1 + r2= 4
C1C2 = $$\sqrt{(2+2)^{2}+(5-3)^{2}}$$
= $$\sqrt{16+4}$$ = $$\sqrt{20}$$
‘C’ divides C1C2 in the ratio 5 : 3

Equation of the pair transverse of the common tangents is
S12 = SS11
(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]
= – (x2 + y2 – 4x – 10y + 28)
⇒ (-x – $$\frac{1}{2}$$y + $$\frac{7}{2}$$)2
= $$\frac{1}{4}$$ (x2 + y2 – 4x – 10y + 28)
(-2x – y + 7)2 = (x2 + y2 – 4x – 10y + 28)
4x2 + y2 + 4xy – 28x – 14y + 49 = x2 + y2 – 4x – 10y + 28
3x2 + 4xy – 24x – 4y + 21 = 0
(3x + 4y – 21); (x – 1) = 0
3x + 4y – 21 = 0; x – 1 = 0

Question 25.
Show that the circles x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 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
x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0
Centres are C1(2, 3), C2 = (-3, -9)
r1 = $$\sqrt{4+9+12}$$ = 5
r2 = $$\sqrt{9+81-26}$$ = 8
C1C2 = $$\sqrt{(2+3)^{2}+(3+9)^{2}}$$
= $$\sqrt{25+144}$$ = 13 = r1 + r2
∴ Circle touch externally .
Equation of common tangent is S1 – S2 = 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 = r2
(x – 1)2 + (y – 4)2 = 52
i.e., x2 + y2– 2x – 8y – 8 = 0

Question 27.
Find the centre and radius of the circle x2 + y2 + 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 3x2 + 3y2 – 6x + 4y – 4 = 0.
Solution:
Given equation is
3x2 + 3y2 – 6x + 4y – 4 = 0
Dividing with 3, we have
x2 + y2 – 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
CP2 = r2 ⇒ r2 = (-1 – 5)2 + (2 – 6)2
= 36 + 16 = 52
Equation of the circle is (x + 1)2 + (y – 2)2
= 52
x2 + 2x + 1 + y2 – 4y + 4 – 52 = 0
x2 + y2 + 2x – 4y – 47 = 0

Question 30.
Find the equation of the circle passing through (2, 3) and concentric with the circle x2 + y2 + 8x + 12y + 15 = 0.
Solution:
The required circle is concentric with the circle x2 + y2 + 8x + 12y + 15 = 0
∴ The equation of the required circle can be taken as
x2 + y2 + 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 x2 + y2 + 8x + 12y – 65 = 0

Question 31.
From the point A(0, 3) on the circle x2 + 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 x2 + y2 + 8x – 6y + 9 = 0, which is a circle.

Question 32.
If the circle x2 + y2 + 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
x2 + y2 + 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 x2 + y2 – 4x + 6y + a = 0 has radius 4 then find a. [A.P. Mar. 16]
Solution:
Equation of the circle is x2 + y2 – 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 x2 + y2 + 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
x2 + y2 – 6x – 8y + 15 = 0

Question 35.
Suppose a point (x1, y1) satisfies x2 + y2+ 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 x2 = coefficient of y2 and coefficient of xy = 0
The given equation represents a circle if g2 + f2 – c ≥ 0
(x1, y1) is a point on the given equation
x2 + y2 + 2gx + 2fy + c = 0, we have
x12 + y12 + 2gx1 + 2fy1 + c = 0
g2 + f2 – c = g2 + f2 + x12 + y12 + 2gx1 + 2fy1 = 0
= (x1 + g)2 + (y1 + 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 (x1, y1) = (1, 2) and (x2, y2) = (4, 5)
Equation of the required circle is
(x – 1) (x – 4) + (y – 2) ( y – 5) = 0
x2 – 5x + 4 + y2 – 7y + 10 = 0
x2 + y2 – 5x – 7y + 14 = 0

Question 37.
Find the other end of the diameter of the circle x2 + y2 – 8x – 8y + 27 = 0 if one end of it is (2, 3).
Solution:
A = (2, 3) and AB is the diameter of the circle x2 + y2 – 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 x2 + y2 + 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
x2 + y2 + $$\frac{c}{a}$$ x + $$\frac{c}{b}$$ y = 0
ab(x2 + y2) + (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 L1 = x + y + 1 = 0; L2 = 3x + y- 5 = 0 and L3 = 2x + y – 5 = 0.
Solution:
Suppose L1, L2,: L2, L3 and L3, L1 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 x2 = Co – efficient of y2
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., x2 + y2 – 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 (x1, y1) = (0, 0); (x2, y1) = (2, 0);
(x3, y3) = (0, 2)
c1 = -(x12 + y12) = 0
c2 = – (x22 + y22) = -(22 + 02) = -4
c3 = -(x32 + y32) = -(02 + 22) – 4
The centre of the circle passing through three non-collinear points P(x1, y1), Q(x2, y2) and R(x3, y3)

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

Question 41.
Obtain the parametric equations of the circle x2 + y2 = 1.
Solution:
Equation of the circle is x2 + y2 = 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
x2 + y2 = 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
x2 + y2 + 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. x2 + y2 – 4x – 6y + 11 = 0.
Solution:
Here (x1, y1) = (2, 4) and
S ≡ x2 + y2 – 4x – 6y + 11
S11 = 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 x2 + y2 – 2x + 4y – 11 = 0.
Solution:
Here (x1, y1) = (1, 3) and
S = x2 + y2 – 2x + 4y – 11 = 0
P(x1, y1) 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
x2 + y2 – 2x + 4y – 20 = 0 ……………… (1)
and x2 + y2 – 2x – 8y + 1 = 0 ……………….. (2)
are in the ratio 2 : 1.
Then show that the equation of the locus of P is x2 + y2 – 2x – 12y + 8 = 0.
Solution:
Let P(x1, y1) 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.
x2 + y2 – 2x + 4y – 20 = 0 and
x2 + y2 – 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 (x12 + y12) – 6x1 – 36y1 + 24 = 0
Locus of P (x1, y1) is
x2 + y2 – 2x – 12y + 8 = 0

Question 46.
If S ≡ x2 + y2 + 2gx + 2fy + c = 0. represents a circle then show that the straight line lx + my + n = 0
i) touches the circle S = 0 if
(g2 + f2 – c) = $$\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}$$

ii) meets the circle S = 0 in two points if
g2 + f2 – c > $$\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}$$

iii) will not meet the circle if
g2 + f2 – 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
g2 + f2 – 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
g2 + f2 – 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
g2 + f2 – 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 x2 + y2 + 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 x2 + y2 – 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 ≡ x2 + y2 + 2gx + 2fy + c = 0 touches the
i) X – axis if g2 = c
ii) Y – axis if f2 = 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 ⇒ g2 = c

Question 50.
Find the equation of the tangent to x2 + y2 – 6x + 4y – 12 = 0 at (- 1, 1).
Solution:
Here (x1, y1) = (-1, 1) and
S ≡ x2 + y2 – 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, y1) to the circle
S ≡ x2 + y2 + 2gx + 2fy + c = 0 is S1 = 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 x2 + y2 – 2x + 4y = 0 at (3, -1). Also find the equation of tangent parallel to it.
Solution:
Here (x1, y1) = (3, -1) and
S ≡ x2 + y2 – 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 x2 + y2 = r2 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 x2 + y2 – 6x + 4y – 12 = 0, then find its point of contact.
Solution:
Let (x1, y1) be the point of contact
Equation of the tangent is
(x1 + g) x + (y1 + f) y + (gx1 + fy1 + 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
3x1 + 4y1 = 1 ………………. (2)
Now by taking first and third equalities of (1), we get
19x1 – 8y1 = -27 ………………. (3)
Solving (2) and (3) we obtain
x1 = -1;, y1 = 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
(x1 ± r cos θ, y1 ± 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., x2 + y2 + 2x – 8y + 4 = 0
and x2 + y2 – 6x + 4y = 0

Question 54.
Show that the line 5x + 12y – 4 = 0 touches the circle x2 + y2 – 6x + 4y + 12 = 0.
Solution:
Equation of the circle is
x2 + y2 – 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 x2 + y2 – 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 θ1 = 30°, θ2 = 60°
Equation of chord joining the point; (-g+ r cos θ1(-f + r sin θ1) where r is the radius of the circle; θ2 and (-g + r cos θ2, -f + r sin θ2) 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 $\frac{60^{\circ}+30^{\circ}}{2}$
(y + 2) sin $\frac{60^{\circ}+30^{\circ}}{2}$
= 5 cos $\frac{60^{\circ}-30^{\circ}}{2}$
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 x2 + y2 + 4x + 6y – 39 = 0.
Solution:
Equation of the circle is
x2 + y2 + 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(x1, y1) to the circle x2 + y2 = a2 with the co-ordinate axes where x1y1 ≠ 0.
Solution:
Equation of the circle is x2 + y2 = a2
Equation of the tangent at P(x1, y1) is xx1 + yy1 = a2 ……………… (1)

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

Question 58.
Find the equation of the normal to the circle x2 + y2 – 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
x1 = 3 and y1 = 2
The equation of the normal at P(x1, y1) of the circle
S ≡ x2 + y2 + 2gx + 2fy + c = Q is
(x – x1) (y1 + f) – (y – y1) (x1 + 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 x2 + y2 – 22x – 4y + 25 = 0 with the co-ordinate axes.
Solution:
Here 2g = -22, 2f = – 4, g = -11, f = -2
x1 = 3, y1 = – 4
Equation of the normal at (3, -4) is
(x – x1) (y1 + f) – (y – y1) (x1 + 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 = x2 + y2 + 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 ≡ x2 + y2 + 2gx + 2fy + c = 0 are perpendicular to each other.
Solution:
If the angle between the tangents drawn from P(x1, y1) to S = 0 is θ, then
tan ($$\frac{\theta}{2}$$) = $$\frac{\mathrm{r}}{\sqrt{\mathrm{S}_{11}}}$$
Equation of the circle is
x2 + y2 + 2gx + 2fy + c = 0
r = $$\sqrt{g^{2}+f^{2}-c}$$
S11 = g2 + f2 + 2g2 + 2f2 + c
= 3g2 + 3f2 + 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 = $$\frac{g^{2}+f^{2}-c}{3 g^{2}+3 f^{2}+c}$$
⇒ 3g2 + 3f2 + c = g2 + f2 – c
⇒ 2g2 + 2f2 + 2c = 0 ⇒ g2 + f2 + 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 θ1, θ2 are the angles of inclination of tangents through a point P to the circle x2 + y2 = a2 then find the locus of P when cot θ1 + cot θ2 = k.
Solution:
Equation of the circle is x2 + y2 = a2
If m is the slope of the tangent, then the equation of tangent passing through P(x1, y1) can be taken as
y1 = mx1 ± a$$\sqrt{1+m^{2}}$$
(y1 – mx1)2 = a2 (1 + m2)
m2x12 + y12 – 2mx1y1 – a2 – a2m2 = 0
m2 (x12 – a2) – 2mx1y1 + (y12 – a2) = 0
Suppose m1 and m2 are the roots of this equation
m1 + m2 = tan θ1 + tan θ2 = $$\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}$$
m1m2 – tan θ1 . tan θ1 = $$\frac{y_{1}^{2}-a^{2}}{x_{1}^{2}-a^{2}}$$
Given that cot θ1 + cot θ2 = 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
2x1y1 = k (y12 – a2)
Locus of P(x1, y1) is 2xy = k(y2 – a2)
Also, conversely if P(x1, y1) satisfies the condition 2xy = k(y2 – a2) then it can be show that cot θ1 + cot θ2 = k
Thus the locus of P is 2xy = k(y2 – a2)

Question 63.
Find the chord of contact of (2, 5) with respect to the circle x2 + y2 – 5x + 4y – 2 = 0.
Solution:
Here (x1, y1) = (2, 5). By
S ≡ x2 + y2 + 2gx + 2fy + c = 0 then the equation of the chord of contact of P with respect to S = 0 is S1 =0, the required chord of contact is
xx1 + yy1 – $$\frac{5}{2}$$ (x + x1) + 2(y + y1) – 2 = 0
Substituting x1 and y1 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 x2 + y2 = a2 cut the circle at A and B such that, AÔB = 90° then show that P lies on the circle x2 + y2 = 2a2.
Solution:
Given circle x2 + y2 = a2 …………… (1)
Let P(x1, y1) 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(x1, y1) with respect to (1) is
xx1 + yy1 – a2 = 0 ……………………. (2)
The equation of the pair of the lines $$\overleftrightarrow{\mathrm{OA}}$$ and $$\overleftrightarrow{\mathrm{OB}}$$ is.given by x2 + y2 – a2
$$\left(\frac{x x_{1}+y y_{1}}{a^{2}}\right)^{2}$$ = 0
or a2 (x2 + y2)- (xx1 + yy1)2 = 0
or x2 (a2 – x12) – 2x1y1xy + y2 (a2 – y12) = 0 – (3)
Since AÔB = 90°, we have the coefficient of x2 in (3) + coefficient of y2 in (3) = 0
∴ a2 – x12 + a2 – y12 = 0
i.e., x12 + y12 = 2a2
Hence the point P(x1, y1) lies on the circle x2 + y2 = 2a2.

Question 65.
Find the equation of the polar of (2, 3) with respect to the circle x2 + y2 + 6x + 8y – 96 = 0.
Solution:
Here (x1, y1) = (2, 3) ⇒ x1 = 2, y1 = 3
Equation of the circle is
x2 + y2 + 6x + 8y – 96 = 0
Equation of the polar is S1 = 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 x2 + y2 – 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 ≡ x2 + y2 – 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 x2 + y2 = a2 with respect to the circle (x + a)2 + y2 = 2a2 lie on y2 + 4ax = 0.
Solution:
Equations of the given circles are
x2 + y2 = a2 …………………. (1)
and (x + a)2 + y2 = 2a2 ………………. (2)
Let P(x1, y1) 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
xx1 + yy1 + a(x + x1) – a2 = 0
x(x1 + a) + yy1 + (ax1 – a2) = 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
(x1 + a)2 + y12 = (x1 – a)2
(or) y12 + (x1 + a)2 – (x1 – a)2 = 0
i.e., y12 + 4ax1 = 0
The poles of the tangents to circle (1) w.r.to circle (2) lie on the curve y2 + 4ax = 0

Question 68.
Show that (4, -2) and (3, -6) are conjugate with respect to the circle x2 + y2 – 24 = 0.
Solution:
Here (x1, y1) = (4, -2) and (x2, y2) = (3, -6) and
S ≡ x2 + y2 – 24 = 0
Two points (x1, y1) and (x2, y2) are conjugate with respect to the circle S = 0 if S12 = 0;
In this case x1x2 + y1y2 – 24 = 0
S12 = 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 x2 + y2 = 17 then find k.
Solution:
Here (x1, y1) = (4, k) and (x2, y2) = (2, 3) and
S ≡ x2 + y2 – 17 = 0
The given points are conjugate ⇒ S12 = 0
x1x2 + y1y2 – 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 x2 + y2 + 4x + 6y – 12 = 0.
Solution:
Here l1 = 2, m1 = 3, n1 = 11
l2 = 2, m2 = -2, n2 = -1
and g = 2, f = 3, c = 12
r = $$\sqrt{9+4-12}$$ = 1
We know that l1x + m1y + n1 = 0
l2x + m2y + n2 = 0 are conjugate with respect to S = 0
r2 (l1l2 + m1m2) = (l1g + m1f – n1) (l2g + m2f – n2)
r2 (l1l2 + m1m2) – 1(2.2 + 3(- 2)) = 4 – 6 = -2
(l1g + m1f – n1) (l2g + m2f – n2)
= (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 ≡ x2 + y2 + 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
x2 + y2 – 2x – 10y + 1 = 0 ………………. (1)
on the line x – 2y + 7 = 0. ……………. (2)
Solution:
Let P(x1, y1) be the mid point of the chord AB
Equation of the chord is S1 = S11
xx1 + yy1 – 1 (x + x1) – 5(y + y1) + 1 = x12 + y12 – 2x1 – 10y1 + 1 .
x(x1 – 1) + y(y1 – 5) – (x12 + y12 – x1 – 5y1) = 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)
x1 – 1 = k ⇒ x1 = k + 1
y1 – 5 = -2(k) ⇒ y1 = 5 – 2k
x12 + y12 – x1 – 5y1 = -7k
⇒ (k + 1)2 + (5 – 2k)2 – (k + 1) – 5(5 – 2k) + 7k = 0
⇒ k2 + 2k + 1 + 25 + 4k2 – 20k – k – 1 – 25 + 10k + 7k = 0
⇒ 5k2 – 2k = 0
⇒ k (5k — 2) = 0 ⇒ k = 0, $$\frac{2}{5}$$
k = 0 ⇒ (x1, y1) = (1, 5) and x – 2y + 7
= 1 – 10 + 7 ≠ 0
(1, 5) is not a point on the chord.
k = $$\frac{2}{5}$$ (x1, y1) = $$\left(\frac{7}{5}, \frac{21}{5}\right)$$

Question 73.
Find the locus of mid-points of the chords of contact of x2 + y2 = a2 from the points lying on the line lx + my + n = 0.
Solution:
Let P = (x1, y1) be a point on the locus P is the mid point of the chord of the circle
x2 + y2 = a2
Equation of the chord is lx + my + n = 0 ……………… (1)
Equation of the circle is x2 + y2 = a2
Equation is the chord having (x1, y1) as mid point of S1 = S11
xx1 + yy1 = x12 + y12
xx1 + yy1 – (x12 + y12) = 0 ……………….. (2)
Pole of (2) with respect to the circle in

Locus of P(x1, y1) is n(x2 + y2) + a2(lx + my) = 0

Question 74.
Show that the four common tangents can be drawn for the circles given by
x2 + y2 – 14x + 6y + 33 = 0 …………… (1)
and x2 + y2 + 30x – 2y +1 = 0 ……………… (2)
and find the internal and external centres of similitude. [T.S. Mar. 19]
Solution:
Equations of the circles are
x2 + y2 – 14x + 6y + 33 = 0
and x2 + y2 + 30x – 2y + 1 = 0
Centres are A(7, -3), B(-15, 1)
r1 = $$\sqrt{49+9-33}$$ = 5
r2 = $$\sqrt{225+1-1}$$ = 15
AB = $$\sqrt{(7+15)^{2}+(-3-1)^{2}}$$
= $$\sqrt{484+16}$$ = $$\sqrt{500}$$ > r1 + r2
∴ Four common tangents can be drawn to the given circle
r1 : r2 = 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 x2 + y2 – 8x – 6y + 21 = 0 and x2 + y2 – 2y – 15 = 0 have exactly two common tangents. Also find the point of intersection of those tangents.
Solution:
Let C1, C2 be the centres and r1, r2 be their radii.
Equation of the circles are
x2 + y2 – 8x – 6y + 21 = 0
and x2 + y2 – 2y – 15 = 0
C1(4, 3), C2(0, 1)
r1 = $$\sqrt{16+9-21}$$ = 2, r2 = $$\sqrt{1+15}$$ = 4
$$\overline{\mathrm{C}_{1} \mathrm{C}_{2}^{2}}$$ = (4 – 0)2 + (3 – 1)2 = 16 + 4 = 20
C1C2 = 2$$\sqrt{5}$$
|r1 – r2| = |2 – 4| = 2, r1 + r2 = 2 + 4 = 6
|r1 – r2| < C1C2 < r1 + r2
Given circles intersect and have exactly two common tangents.
r1 : r2 = 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 x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 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
x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0
Centres are C1(2, 3), C2 = (-3, -9)
r1 = $$\sqrt{4+9+12}$$ = 5
r2 = $$\sqrt{9+81-26}$$ = 8
C1C2 = $$\sqrt{(2+3)^{2}+(3+9)^{2}}$$
= $$\sqrt{25+144}$$ = 13 = r1 + r2
∴ Circle touch externally
Equation of common tangent is S1 – S2 = 0
-10x – 24y – 38 = 0
5x + 12y + 19 = 0

Question 77.
Show that the circles x2 + y2 – 4x – 6y – 12 = 0 and 5(x2 + y2) – 8x – 14y – 32 = 0 touch each other and find their point of contact.
Solution:
Equations of the circles are
x2 + y2 – 4x – 6y – 12 = 0
and x2 + y2 – $$\frac{8}{5}$$ 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 x2 + y2 = 25.
Solution:
(x1, y1) = (1o, 4)
Equation of the circle is x2 + y2 – 25 = 0
Equation of the pair of tangents is S12 = S . S11
(10x + 4y – 25)2 = (100 + 16 – 25)(x2 + y2 – 25)
100x2 + 16y2 + 625 + 80xy – 500x – 200y = 91x2 + 91y2 – 2275
9x2 + 80xy – 75y2 – 500x – 200y + 2900 = 0

Question 79.
Find the equation to all possible common tangents of the circles x2 + y2 – 2x – 6y + 6 = 0 and x2 + y2 = 1.
Solution:
Equations of the circles are
x2 + y2 – 2x – 6y + 6 = 0
and x2 + y2 = 1
Centres are A(1, 3), B(0, 0),
r1 = $$\sqrt{1+9-6}$$ = 2
r2 = 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
(x2 + y2 – 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) (x2 + y2 – 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