If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = 4$ orthogonally, then the locus of its centre is $m a x + m a y - (a^{2} + b^{2} + p^{2})$ .Find $m$

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Solution

Let equation of circle be :
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0. F o r o r t h o g o n a l c i r c l e s : 2 g_{1} g_{2} + 2 f_{1} f_{2} = C_{1} + C_{2} \Rightarrow 0 + 0 = C_{1} - 4 = C - 4 = 0 \Rightarrow C = 4. a n d a^{2} + b^{2} + 2 a g + 2 b f + c = 0. c e n t r e \to (- g, - f) ∴ h = - g & k = - f {\Rightarrow a}^{2} + b^{2} - 2 a h - 2 b k + 4 = 0. \Rightarrow L o c u s : {\Rightarrow a}^{2} + b^{2} - 2 a x - 2 b y + 4 = 0. 2 a x + 2 b y - (a^{2} + b^{2} + 4) = 0. ∴ m = 2, p = 2.$

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