Find the equation of the circle whose centre is at the point (4,5) and which passes through the centre of the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ .

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Solution

$\begin{matrix} x^{2} + y^{2} - 6 x + y - 12 = 0 (x^{2} - 6 x + 9) + (y^{2} + 4 y + 4) = 25 {(x - 3)}^{2} + {(y + 2)}^{2} = 5^{2} C e n t r e (3, - 2), r_{1} = 5 S o, e q u a t i o n o f c i r c l e {(x - 4)}^{2} + {(y - 5)}^{2} = {(4 - 3)}^{2} + {(5 + 2)}^{2} {(x - 4)}^{2} + {(y - 5)}^{2} = 50 \end{matrix}$

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