# Long Questions

In the given figure, the circle A with centre X passes through the centre Y of the circle B. If the equation of circle B is x^{2} + y^{2} - 4x + 6y - 12 = 0 and the co-ordinates of X are (-4, 5), then find the equation of the circle A.

Comparing the equation of circle B i.e. x

^{2}+ y

^{2}- 4x + 6y - 12 = 0 with x

^{2}+ y

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

g = -2, f = 3

So, the center of the circle B is Y(h, k) = Y(-g, -f) = Y(2, -3)

Now, as shown in figure, circle A having centre X(-4, 5) passes through the centre of circle B Y(2, -3), so,
radius (r) = $\sqrt{(x_2 \text{ - }x_1)^2 + (y_2 \text{ - }y_1)^2}$
or, r^{2} = $(\sqrt{(\text{ 2 + 4 })^2 + (\text{ -3 - 5})^2})^2$

or, r^{2} = 100 --- (i)

^{2}+ (y - k)

^{2}= r

^{2}or, (x + 4)

^{2}+ (y - 5)

^{2}= 100 (from 'i')

or, x

^{2}+ 8x + 16 + y

^{2}- 10y + 25 = 100

so, x

^{2}+ y

^{2}+ 8x - 10y - 59 = 0 is the required equation.

Find the equation of circle with centre (-1, 2) and passing through the centre of the circle having equation
x^{2} + y^{2} - 6x - 10y - 2 = 0.

Comparing the equation of given circle i.e. x

^{2}+ y

^{2}- 6x - 10y - 2 = 0 with x

^{2}+ y

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

g = -3, f = 5

so, the center of the given circle is (h, k) = (-g, -f) = (3, 5)

Now, as given, required circle having centre (-1, 2) passes through the centre of given circle (3, 5), so,
radius (r) = $\sqrt{(x_2 \text{ - }x_1)^2 + (y_2 \text{ - }y_1)^2}$
or, r^{2} = $(\sqrt{(\text{ 3 + 1 })^2 + (\text{ 5 - 2})^2})^2$

or, r^{2} = 25 --- (i)

Finally, the required equation of circle having center (-1, 2) is given by,
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x + 1)^{2} + (y - 2)^{2} = 25 (from 'i')

or, x^{2} + 2x + 1 + y^{2} - 4y + 4 = 25

so, x^{2} + y^{2} + 2x - 4y - 20 = 0 is the required equation.

If the coordinate of one end of a diameter of the circle having equation
x^{2} + y^{2} + 4x - 6y + 8 = 0 is (-4, 4), then find the
coordinates of the other end of the diameter.

Comparing the given equation of circle, x^{2} + y^{2} + 4x - 6y + 8 = 0 with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get,

g = 2, f = -3

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

so, -2 = $\frac{\text{-4 + a}}{2}$ and 3 = $\frac{\text{4 + b}}{2}$

so, a = 0 and b = 2

Hence, another coordinate of the given circle is (a, b) = (0, 2).

Find the equation of the circle passing through the points (2, 3) and (5, 4) and centre on the line 2x + 3y - 7 = 0.

(x - h)

^{2}+ (y - k)

^{2}= r

^{2}--- (A)

Center of circle lies on the line 2x + 3y - 7 = 0. so,
2h + 3k - 7 = 0

or, h = $\frac{\text{7 - 3k}}{2}$ --- (i)

It pass through the point (2, 3), so from (i),
(2 - h)^{2} + (3 - k)^{2} = r^{2} --- (ii)

It also pass through the point (5, 4), so from (i),
(5 - h)^{2} + (4 - k)^{2} = r^{2} --- (iii)

From (ii) and (iii),
(2 - h)^{2} + (3 - k)^{2} = (5 - h)^{2} + (4 - k)^{2}

or, 4 - 4h + $\cancel{h^2}$ + 9 - 6k + $\cancel{k^2}$ = 25 - 10h + $\cancel{h^2}$ + 16 - 8k + $\cancel{k^2}$

or, -28 + 6h + 2k = 0

or, -28 + 6$(\frac{\text{7 - 3k}}{2})$ + 2k = 0 (using 'i')

or, -28 + 21 - 9k + 2k = 0

or, 7k = -7

so, k = -1 --- (iv)

and, h = 5 (from 'i') --- (v)

Now, from (ii),
(2 - 5)^{2} + (3 + 1)^{2} = r^{2} (using 'iv' and 'v')
or, 9 + 16 = r^{2}

so, r^{2} = 25 --- (vi)

Finally, using (iv), (v) and (vi) in equation (A),
(x - 5)^{2} + (y + 1)^{2} = 25
or, x^{2} - 10x + 25 + y^{2} + 2y + 1 = 25

or, x^{2} + y^{2} - 10x + 2y + 1 = 0

Hence, x^{2} + y^{2} - 10x + 2y + 1 = 0 is the required equation of circle.

Find the equation of the circle having the centre (4, 6) and passing through the midpoint of the line joining the points (-1, 3) and (3, 1).

^{2}+ (y - k)

^{2}= r

^{2}or, (x - 4)

^{2}+ (y - 6)

^{2}= r

^{2}--- (i)

Now, midpoint of the line joining th points (-1, 3) and (3, 1) is, (x, y) = $(\frac{\text{-1 + 3}}{2}, \frac{\text{3 + 1}}{2})$ = (1, 2)

Since, required circle pass through (1, 2), from (i),
(1 - 4)^{2} + (2 - 6)^{2} = r^{2}
so, r^{2} = 25 --- (ii)

Finally, using (ii) in (i),
(x - 4)^{2} + (y - 6)^{2} = 25
or, x^{2} - 8x + 16 + y^{2} - 12y + 36 = 25

or, x^{2} + y^{2} - 8x - 12y + 27 = 0

Hence, x^{2} + y^{2} - 8x - 12y + 27 = 0 is the required equation of circle.

Find the equation of the circle having centre (1, 2) and passing through the point of intersection of the lines x + 2y = 3 and 3x + y = 4.

^{2}+ (y - k)

^{2}= r

^{2}or, (x - 1)

^{2}+ (y - 2)

^{2}= r

^{2}--- (i)

Now, Given lines,
x + 2y = 3
so, x = 3 - 2y --- (ii)

and, 3x + y = 4 --- (iii)

Using (ii) in (iii),
3(3 - 2y) + y = 4
or, 9 - 6y + y = 4

or, 5 = 5y

so, y = 1

and, x = 1

Since, required circle pass through (1, 1), from (i),
(1 - 1)^{2} + (1 - 2)^{2} = r^{2}
or, r^{2} = 1 --- (iv)

Finally, from (i) and (iv),
(x - 1)^{2} + (y - 2)^{2} = 1
or, x^{2} - 2x + 1 + y^{2} - 4y + 4 = 1

or, x^{2} + y^{2} - 2x - 4y + 4 = 0

Hence, x^{2} + y^{2} - 2x - 4y + 4 = 0 is the required equation of circle.

Find the equation of a circle concentric with the circle x^{2} + y^{2}
- 6x + y = 1 and passing through the point (4, -2).

^{2}+ y

^{2}- 6x + y = 1 --- (i)

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

g = -3, f = ½

So, center = (h, k) = (-g, -f) = (3, -½).

Since, required circle and (i) is concentric, center (h, k) of required circle is also (3, -½).

so, we have,
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x - 3)^{2} + (y + ½)^{2} = r^{2} --- (ii)

Again, required circle pass through (4, -2). so, distance between center and (4, -2) is,
r^{2} = $(\sqrt{(x_2 \text{ - }x_1)^2 + (y_2 \text{ - }y_1)^2})^2$
or, r^{2} = (4 - 3)^{2} + (-2 + ½)^{2}

so, r^{2} = $\frac{13}{4}$ --- (iii)

Finally, using (iii) in (ii),
(x - 3)^{2} + (y + ½)^{2} = $\frac{13}{4}$
or, x^{2} + y^{2} - 6x + y + 6 = 0

Hence, x^{2} + y^{2} - 6x + y + 6 = 0 is the required equation of a circle.

Find the equation of the circle which passes through the points (3, 2) and (5, 4) and its centre lies on the line 3x - 2y - 1 = 0.

(x - h)

^{2}+ (y - k)

^{2}= r

^{2}--- (A)

Center of circle lies on the line 3x - 2y - 1 = 0. so,
3h - 2k - 1 = 0

or, h = $\frac{\text{2k + 1}}{3}$ --- (i)

It pass through the point (3, 2), so from (i),

(3 - h)^{2} + (2 - k)^{2} = r^{2} --- (ii)

It also pass through the point (5, 4), so from (i),

(5 - h)^{2} + (4 - k)^{2} = r^{2} --- (iii)

From (ii) and (iii),
(3 - h)^{2} + (2 - k)^{2} = (5 - h)^{2} + (4 - k)^{2}
or, 9 - 6h + $\cancel{h^2}$ + 4 - 4k + $\cancel{k^2}$ = 25 - 10h + $\cancel{h^2}$ + 16 - 8k + $\cancel{k^2}$

or, 13 - 6h - 4k = 41 - 10h - 8k

or, 4h = 28 - 4k

or, 4$(\frac{\text{2k + 1}}{3})$ = 28 - 4k

or, 8k + 4 = 84 - 12k

or, 20k = 80

so, k = 4

and, h = 3 (from 'i')

Now, from (ii),
(3 - 3)^{2} + (2 - 4)^{2} = r^{2} (using 'iv' and 'v')
so, r^{2} = 4 --- (vi)

Finally, using (iv), (v) and (vi) in equation (A),
(x - 3)^{2} + (y - 4)^{2} = 4
or, x^{2} - 6x + 9 + y^{2} - 8y + 16 = 4

or, x^{2} + y^{2} - 6x - 8y + 21 = 0

Hence, x^{2} + y^{2} - 6x - 8y + 21 = 0 is the required equation of circle.

Find the equation of the circle whose centre lies on the line x - 4y = 1 and which passes through the points (3, 7) and (5, 5).

(x - h)

^{2}+ (y - k)

^{2}= r

^{2}--- (A)

Center of circle lies on the line x - 4y = 1. so,
h - 4k = 1

or, h = 1 + 4k --- (i)

It pass through the point (3, 7), so from (i),

(3 - h)^{2} + (7 - k)^{2} = r^{2} --- (ii)

It also pass through the point (5, 5), so from (i),

(5 - h)^{2} + (5 - k)^{2} = r^{2} --- (iii)

From (ii) and (iii),
(3 - h)^{2} + (7 - k)^{2} = (5 - h)^{2} + (5 - k)^{2}
or, 9 - 6h + $\cancel{h^2}$ + 49 - 14k + $\cancel{k^2}$ = 25 - 10h + $\cancel{h^2}$ + 25 - 10k + $\cancel{k^2}$

or, 58 - 6h - 14k = 50 - 10h - 10k

or, 8 + 4h = 4k

or, 8 + 4(1 + 4k) = 4k (using 'i')

or, 8 + 4 + 16k = 4k

or, 12 = -12k

so, k = -1

amd, h = -3 (from 'i')

Now, from (ii),
(3 + 3)^{2} + (7 + 1)^{2} = r^{2} (using 'iv' and 'v')
so, r^{2} = 100 --- (vi)

Finally, using (iv), (v) and (vi) in equation (A),
(x + 3)^{2} + (y + 1)^{2} = 100
or, x^{2} + 6x + 9 + y^{2} 2y + 1 = 100

or, x^{2} + y^{2} + 6x + 2y - 90 = 0

Hence, x^{2} + y^{2} + 6x + 2y - 90 = 0 is the required equation of circle.

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

x

^{2}+ y

^{2}- 6x + y = 1 --- (i)

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

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

So, center of given circle = (h', k') = (-g, -f) = (1, -2).

Since, required circle having center (h, k) = (3, 2) pass through the center of given circle (h', k') = (1, -2),
r^{2} = $(\sqrt{(h' \text{ - }h)^2 + (k' \text{ - }k)^2})^2$
or, r^{2} = (1 - 3)^{2} + (-2 - 2)^{2}

so, r^{2} = 20 units --- (iii)

Finally, equation of circle having centre (3, 2) is give as,
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x - 3)^{2} + (y - 2)^{2} = 20

or, x^{2} - 6x + 9 + y^{2} - 4y + 4 = 20

or, x^{2} + y^{2} - 6x - 4y - 7 = 0

Hence, x^{2} + y^{2} - 6x - 4y - 7 = 0 is the required equation of a circle.

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

^{2}+ 2y

^{2}+ 4x - 2y + 1 = 0 or, x

^{2}+ y

^{2}+ 2x - y + ½ = 0 --- (i)

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

g = 1, f = -½

So, center = (h, k) = (-g, -f) = (-1, ½).

Since, required circle and (i) is concentric, center (h, k) of required circle is also (-1, ½).

so, we have,
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x + 1)^{2} + (y - ½)^{2} = r^{2} --- (ii)

Again, required circle pass through (4, -2). so, distance between center (-1, ½) and (4, -2) is,
r^{2} = $(\sqrt{(x_2 \text{ - }x_1)^2 + (y_2 \text{ - }y_1)^2})^2$
or, r^{2} = (4 + 1)^{2} + (-2 - ½)^{2}

or, r^{2} = 25 + $\frac{25}{4}$

so, r^{2} = $\frac{125}{4}$ --- (iii)

Finally, using (iii) in (ii),
(x + 1)^{2} + (y - ½)^{2} = $\frac{125}{4}$
or, x^{2} + 2x + 1 + y^{2} - y + $\frac14$ - $\frac{125}{4}$ = 0

or, x^{2} + y^{2} + 2x - y - 30 = 0

Hence, x^{2} + y^{2} + 2x - y - 30 = 0 is the required equation of a circle.

Find the equation of a circle passing through the point (4, 3) and concentric
with the circle having equation x^{2} + y^{2} + 6x - 8y - 11 =
0.

x

^{2}+ y

^{2}+ 6x - 8y - 11 = 0 --- (i)

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

g = 3, f = -4

So, center = (h, k) = (-g, -f) = (-3, 4).

Since, required circle and (i) is concentric, center (h, k) of required circle is also (-3, 4).

so, we have,
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x + 3)^{2} + (y - 4)^{2} = r^{2} --- (ii)

Again, required circle pass through (4, 3). so, distance between center (-3, 4) and (4, 3) is,
r^{2} = $(\sqrt{(x_2 \text{ - }x_1)^2 + (y_2 \text{ - }y_1)^2})^2$
or, r^{2} = (4 + 3)^{2} + (3 - 4)^{2}

or, r^{2} = 50 units --- (iii)

Finally, using (iii) in (ii),
(x + 3)^{2} + (y - 4)^{2} = 50
or, x^{2} + 6x + 9 + y^{2} - 8y + 16 - 50 = 0

or, x^{2} + y^{2} + 6x - 8y -25 = 0

Hence, x^{2} + y^{2} + 6x - 8y -25 = 0 is the required equation of a circle.

Find the centre and the length of the diameter of a circle having equation
4x^{2} + 4y^{2} - 24x - 20y - 3 = 0.

^{2}+ 4y

^{2}- 24x - 20y - 3 = 0 or, x

^{2}+ y

^{2}- 6x - 5y - ¾ = 0

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

g = -3, f = -$\frac52$, c = - ¾

So, center of circle is (h, k) = (-g, -f) = (3, $\frac52$).

And, radius (r) of circle is given by,

r = $\sqrt{g^2 + f^2 - c}$
= $\sqrt{\text{9 + }\frac{25}{4} \text{ + } \frac 34}$

= 4 units

So, center of circle is (h, k) = (3, $\frac52$) and diameter = 2 × r = 8 units.

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

(x - h)

^{2}+ (y - k)

^{2}= r

^{2}--- (A)

Center of circle lies on the line 4x + y = 16. so,
4h + k = 16

or, k = 16 - 4h --- (i)

It pass through the point (4, 1), so from (i),

(4 - h)^{2} + (1 - k)^{2} = r^{2} --- (ii)

It also pass through the point (6, 5), so from (i),

(6 - h)^{2} + (5 - k)^{2} = r^{2} --- (iii)

From (ii) and (iii),
(4 - h)^{2} + (1 - k)^{2} = (6 - h)^{2} + (5 - k)^{2}
or, 16 - 8h + $\cancel{h^2}$ + 1 - 2k + $\cancel{k^2}$ = 36 - 12h + $\cancel{h^2}$ + 25 - 10k + $\cancel{k^2}$

or, 17 - 8h - 2k = 61 - 12h - 10k

or, 4h = 4 - 8k

or, 4h = 44 - 6(16 - 4h) (using 'i')

or, 4h = 44 - 128 + 32h

or, 84 = 28h

so, h = 3

and, k = 4 (from 'i')

Now, from (ii),
(4 - 3)^{2} + (1 - 4)^{2} = r^{2} (using 'iv' and 'v')
or,1 + 9 = r^{2}

so, r^{2} = 10 --- (vi)

Finally, using (iv), (v) and (vi) in equation (A),
(x - 3)^{2} + (y - 4)^{2} = 10
or, x^{2} - 6x + 9 + y^{2} - 8y + 16 = 10

or, x^{2} + y^{2} - 6y - 8y + 15 = 0

Hence, x^{2} + y^{2} - 6y - 8y + 15 = 0 is the required equation of circle.

Find the equation of the circle which passes through the points P(5, 7), Q(6, 6) and R(2, -2).

(x - h)

^{2}+ (y - k)

^{2}= r

^{2}--- (A)

It pass through P(5, 7), so,

(5 - h)^{2} + (7 - k)^{2} = r^{2} --- (i)

It also pass through Q(6, 6), so,

(6 - h)^{2} + (6 - k)^{2} = r^{2} --- (ii)

And, it pass through Q(2, -2), so,
(2 - h)^{2} + (-2 - k)^{2} = r^{2}
or, (2 - h)^{2} + {-(2 + k)}^{2} = r^{2}

or, (2 - h)^{2} + (2 + k)^{2} = r^{2} --- (iii)

From (i) and (ii),
(5 - h)^{2} + (7 - k)^{2} = (6 - h)^{2} + (6 - k)^{2}
or, 25 - 10h + $\cancel{h^2}$ + 49 - 14k + $\cancel{k^2}$ = 36 - 12h + $\cancel{h^2}$ + 36 - 12k + $\cancel{k^2}$

or, 74 - 10h - 14k = 72 - 12h - 12k

or, 2h = -2 + 2k

or, h = k - 1 --- (iv)

Also, taking (ii) and (iii),
(6 - h)^{2} + (6 - k)^{2} = (2 - h)^{2} + (2 + k)^{2}
or, 36 - 12h + $\cancel{h^2}$ + 36 - 12k + $\cancel{k^2}$ = 4 - 4h + $\cancel{h^2}$ + 4 + 4k + $\cancel{k^2}$

or, 72 - 12h - 12k = 8 - 4h + 4k

or, 64 = 8h + 16k

or, 64 = 8k - 8 + 16k (using 'iv')

or, k = 3 --- (v)

and, h = 2 (from 'iv') --- (vi)

Now, using (v) and (vi) in (iii),
(2 - 2)^{2} + (2 + 3)^{2} = r^{2}
so, r^{2} = 25 --- (vii)

Finally, using (v), (vi) and (vii) in (A),
(x - 2)^{2} + (y - 3)^{2} = 25
or, x^{2} - 4x + 4 + y^{2} - 6y + 9 = 25

or, x^{2} + y^{2} - 4x - 6y - 12 = 0

^{2}+ y

^{2}- 4x - 6y - 12 = 0 is the required equation of circle.