If a circle passes through the point (a , b) | vedclass

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Question

Let the circle be $x^{2}+y^{2}+2 g x+2 f y+c=0$

since it cuts $x^{2}+y^{2}-k^{2}=0$ orthogonally therefore

$2 g(0)+2 f(0)=c-k^{2}=0 	herefore c

Vedclass · Accepted Answer

$2ax + 2by - (a^2 + b^2 + K^2) = 0$

Vedclass · Answer

Let the circle be $x^{2}+y^{2}+2 g x+2 f y+c=0$

since it cuts $x^{2}+y^{2}-k^{2}=0$ orthogonally therefore

$2 g(0)+2 f(0)=c-k^{2}=0 	herefore c=k^{2}$

Again the circle $C_{2}$ passes through $(a, b)$ $	herefore a^{2}+b^{2}+2 g a+2 f b+k^{2}=0 \because c=k^{2}$

$	herefore$ Locus of centre $(-g,-f)$ is $a^{2}+b^{2}+2 a(-x)+2 b(-y)+k^{2}=0$

or $2 a x+2 b y-\left(a^{2}+b^{2}+k^{2}ight)=0$

If a circle passes through the point $(a , b) \&$ cuts the circle $x^2 + y^2= K^2$ orthogonally, then the equation of the locus of its centre is :

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