JEE Advance - Mathematics (1988 - No. 24)
If a circle passes through the point (a, b) and cuts the circle $${x^2}\, + \,{y^2}\, = \,{k^2}$$ orthogonally, then the equation of the locus of its centre is
$$2\,ax\, + \,2\,by\, - \,({a^2}\, + \,{b^2}\, + \,\,{k^2})\, = \,0$$
$$2\,ax\, + \,2\,by\, - \,({a^2}\, - \,\,{b^2}\, + \,\,{k^2})\, = \,0$$
$${x^2}\, + \,{y^2}\, - \,3\,\,ax\, + \,4\,by\, + \,\,({a^2}\, + \,\,{b^2}\, - \,\,{k^2})\, = \,0$$
$${x^2}\, + \,{y^2}\, - \,2\,\,ax\, - \,4\,by\, + \,\,({a^2}\, - \,\,{b^2}\, - \,\,{k^2})\, = \,0$$.
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