Long Questions

Here,
Comparing the equation of circle B i.e. x² + y² - 4x + 6y - 12 = 0 with x² + y² + 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, r² = $(\sqrt{(\text{ 2 + 4 })^2 + (\text{ -3 - 5})^2})^2$
or, r² = 100 --- (i)

Finally, the equation of circle A with center X(-4, 5) is given by, (x - h)² + (y - k)² = r² or, (x + 4)² + (y - 5)² = 100 (from 'i')
or, x² + 8x + 16 + y² - 10y + 25 = 100
so, x² + y² + 8x - 10y - 59 = 0 is the required equation.

Here,
Comparing the equation of given circle i.e. x² + y² - 6x - 10y - 2 = 0 with x² + y² + 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, r² = $(\sqrt{(\text{ 3 + 1 })^2 + (\text{ 5 - 2})^2})^2$
or, r² = 25 --- (i)

Finally, the required equation of circle having center (-1, 2) is given by, (x - h)² + (y - k)² = r² or, (x + 1)² + (y - 2)² = 25 (from 'i')
or, x² + 2x + 1 + y² - 4y + 4 = 25
so, x² + y² + 2x - 4y - 20 = 0 is the required equation.

Let (a, b) be the coordinates of another end of a diameter.

Comparing the given equation of circle, x² + y² + 4x - 6y + 8 = 0 with x² + y² + 2gx + 2fy + c = 0, we get,
g = 2, f = -3
So, center = C(h, k) = C(-g, -f) = C(-2, 3)

Now, using midpoint formula, C(h, k) = $(\frac{\text{-4 + a}}{2}, \frac{\text{4 + b}}{2})$ or, C(-2, 3) = $(\frac{\text{-4 + a}}{2}, \frac{\text{4 + b}}{2})$
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).

Here, Equation of circle is given by,
(x - h)² + (y - k)² = r² --- (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 = r² --- (ii)

It also pass through the point (5, 4), so from (i), (5 - h)² + (4 - k)² = r² --- (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 = r² (using 'iv' and 'v') or, 9 + 16 = r²
so, r² = 25 --- (vi)

Finally, using (iv), (v) and (vi) in equation (A), (x - 5)² + (y + 1)² = 25 or, x² - 10x + 25 + y² + 2y + 1 = 25
or, x² + y² - 10x + 2y + 1 = 0

Hence, x² + y² - 10x + 2y + 1 = 0 is the required equation of circle.

Here, Equation of circle having center (4, 6) is given by, (x - h)² + (y - k)² = r² or, (x - 4)² + (y - 6)² = r² --- (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 = r² so, r² = 25 --- (ii)

Finally, using (ii) in (i), (x - 4)² + (y - 6)² = 25 or, x² - 8x + 16 + y² - 12y + 36 = 25
or, x² + y² - 8x - 12y + 27 = 0

Hence, x² + y² - 8x - 12y + 27 = 0 is the required equation of circle.

Here, Equation of circle having center (1, 2) is given by, (x - h)² + (y - k)² = r² or, (x - 1)² + (y - 2)² = r² --- (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 = r² or, r² = 1 --- (iv)

Finally, from (i) and (iv), (x - 1)² + (y - 2)² = 1 or, x² - 2x + 1 + y² - 4y + 4 = 1
or, x² + y² - 2x - 4y + 4 = 0

Hence, x² + y² - 2x - 4y + 4 = 0 is the required equation of circle.

Here, given equation of circle, x² + y² - 6x + y = 1 --- (i)

Comparing it with, x² + y² + 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 = r² or, (x - 3)² + (y + ½)² = r² --- (ii)

Again, required circle pass through (4, -2). so, distance between center and (4, -2) is, r² = $(\sqrt{(x_2 \text{ - }x_1)^2 + (y_2 \text{ - }y_1)^2})^2$ or, r² = (4 - 3)² + (-2 + ½)²
so, r² = $\frac{13}{4}$ --- (iii)

Finally, using (iii) in (ii), (x - 3)² + (y + ½)² = $\frac{13}{4}$ or, x² + y² - 6x + y + 6 = 0

Hence, x² + y² - 6x + y + 6 = 0 is the required equation of a circle.

Here, Equation of circle is given by,
(x - h)² + (y - k)² = r² --- (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 = r² --- (ii)

It also pass through the point (5, 4), so from (i),
(5 - h)² + (4 - k)² = r² --- (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 = r² (using 'iv' and 'v') so, r² = 4 --- (vi)

Finally, using (iv), (v) and (vi) in equation (A), (x - 3)² + (y - 4)² = 4 or, x² - 6x + 9 + y² - 8y + 16 = 4
or, x² + y² - 6x - 8y + 21 = 0

Hence, x² + y² - 6x - 8y + 21 = 0 is the required equation of circle.

Here, Equation of circle is given by,
(x - h)² + (y - k)² = r² --- (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 = r² --- (ii)

It also pass through the point (5, 5), so from (i),
(5 - h)² + (5 - k)² = r² --- (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 = r² (using 'iv' and 'v') so, r² = 100 --- (vi)

Finally, using (iv), (v) and (vi) in equation (A), (x + 3)² + (y + 1)² = 100 or, x² + 6x + 9 + y² 2y + 1 = 100
or, x² + y² + 6x + 2y - 90 = 0

Hence, x² + y² + 6x + 2y - 90 = 0 is the required equation of circle.

Here, given equation of circle,
x² + y² - 6x + y = 1 --- (i)

Comparing it with, x² + y² + 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), r² = $(\sqrt{(h' \text{ - }h)^2 + (k' \text{ - }k)^2})^2$ or, r² = (1 - 3)² + (-2 - 2)²
so, r² = 20 units --- (iii)

Finally, equation of circle having centre (3, 2) is give as, (x - h)² + (y - k)² = r² or, (x - 3)² + (y - 2)² = 20
or, x² - 6x + 9 + y² - 4y + 4 = 20
or, x² + y² - 6x - 4y - 7 = 0

Hence, x² + y² - 6x - 4y - 7 = 0 is the required equation of a circle.

Here, given equation of circle, 2x² + 2y² + 4x - 2y + 1 = 0 or, x² + y² + 2x - y + ½ = 0 --- (i)

Comparing it with, x² + y² + 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 = r² or, (x + 1)² + (y - ½)² = r² --- (ii)

Again, required circle pass through (4, -2). so, distance between center (-1, ½) and (4, -2) is, r² = $(\sqrt{(x_2 \text{ - }x_1)^2 + (y_2 \text{ - }y_1)^2})^2$ or, r² = (4 + 1)² + (-2 - ½)²
or, r² = 25 + $\frac{25}{4}$
so, r² = $\frac{125}{4}$ --- (iii)

Finally, using (iii) in (ii), (x + 1)² + (y - ½)² = $\frac{125}{4}$ or, x² + 2x + 1 + y² - y + $\frac14$ - $\frac{125}{4}$ = 0
or, x² + y² + 2x - y - 30 = 0

Hence, x² + y² + 2x - y - 30 = 0 is the required equation of a circle.

Here, given equation of circle,
x² + y² + 6x - 8y - 11 = 0 --- (i)

Comparing it with, x² + y² + 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 = r² or, (x + 3)² + (y - 4)² = r² --- (ii)

Again, required circle pass through (4, 3). so, distance between center (-3, 4) and (4, 3) is, r² = $(\sqrt{(x_2 \text{ - }x_1)^2 + (y_2 \text{ - }y_1)^2})^2$ or, r² = (4 + 3)² + (3 - 4)²
or, r² = 50 units --- (iii)

Finally, using (iii) in (ii), (x + 3)² + (y - 4)² = 50 or, x² + 6x + 9 + y² - 8y + 16 - 50 = 0
or, x² + y² + 6x - 8y -25 = 0

Hence, x² + y² + 6x - 8y -25 = 0 is the required equation of a circle.

Here, Given equation of circle is, 4x² + 4y² - 24x - 20y - 3 = 0 or, x² + y² - 6x - 5y - ¾ = 0

Comparing it with, x² + y² + 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.

Here, Equation of circle is given by,
(x - h)² + (y - k)² = r² --- (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 = r² --- (ii)

It also pass through the point (6, 5), so from (i),
(6 - h)² + (5 - k)² = r² --- (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 = r² (using 'iv' and 'v') or,1 + 9 = r²
so, r² = 10 --- (vi)

Finally, using (iv), (v) and (vi) in equation (A), (x - 3)² + (y - 4)² = 10 or, x² - 6x + 9 + y² - 8y + 16 = 10
or, x² + y² - 6y - 8y + 15 = 0

Hence, x² + y² - 6y - 8y + 15 = 0 is the required equation of circle.

Here, Equation of circle is given by,
(x - h)² + (y - k)² = r² --- (A)

It pass through P(5, 7), so,
(5 - h)² + (7 - k)² = r² --- (i)

It also pass through Q(6, 6), so,
(6 - h)² + (6 - k)² = r² --- (ii)

And, it pass through Q(2, -2), so, (2 - h)² + (-2 - k)² = r² or, (2 - h)² + {-(2 + k)}² = r²
or, (2 - h)² + (2 + k)² = r² --- (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 = r² so, r² = 25 --- (vii)

Finally, using (v), (vi) and (vii) in (A), (x - 2)² + (y - 3)² = 25 or, x² - 4x + 4 + y² - 6y + 9 = 25
or, x² + y² - 4x - 6y - 12 = 0

Hence, x² + y² - 4x - 6y - 12 = 0 is the required equation of circle.