Question: In the figure, the circle A with centre X passes through the centre Y of the circle B. If the equation of circle B is x² + y² - 4x + 6y - 12 = 0 and the co-ordinates of X are (-4,5), then find the equation of circle A.

Solution:
Given,
Figure of the Question

Equation of circle B is x² +y² -4x +6y -12 = 0
or, x² + y² - 2×2x + 2×3y - 12 = 0 - (i)

Comparing above equation (i) with x² + y² + 2gx + 2fy + c = 0, we get,
-g = 2, -f = -3
So, centre of circle B (Y) is (2,-3)

Since, the point Y is the passing point of circle A, (x,y) = (2,-3)

In circle A, 
Co-ordinates of centre of circle A (h,k) = (-4,5)
When (x,y) = (2,-3) and (h,k) = (-4,5), using formula,
r² = (x - h)² + (y - k)²
or, r² = {2 - (-4)}² + (-3 -5)²
or, r² = 6² + 8²
So, r = 10 units
Equation of circle A when centre (h,k) = (-4,5) and radius (r) = 10, we get,

$(x - h)^2 + (y - k)^2 = r^2$
$or, \{x - (-4)\}^2 + \{y - 5\}^2 = 10^2$

$or, \(x +4)^2 + (y -5)^2 = 100$

$or, x^2 + y^2 + 8x - 10y - 59 = 0$ is the required equation.

Hence, the required equation of the given circle A is x² + y² + 8x - 10y - 59 = 0.

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