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

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(B) Let the equation of the variable circle be $x^{2} + y^{2} + 2gx + 2fy + c = 0$ ... $(i)$.Since the circle $(i)$ intersects the circle $x^{2} + y^{

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$2ax + 2by - (a^{2} + b^{2} + 4) = 0$

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(B) Let the equation of the variable circle be $x^{2} + y^{2} + 2gx + 2fy + c = 0$ ... $(i)$.Since the circle $(i)$ intersects the circle $x^{2} + y^{2} - 4 = 0$ orthogonally,the condition $2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}$ applies.Here,$g_{2} = 0, f_{2} = 0, c_{2} = -4$.So,$2g(0) + 2f(0) = c - 4$,which gives $c = 4$.Since the circle passes through the point $(a, b)$,we have $a^{2} + b^{2} + 2ga + 2fb + 4 = 0$.To find the locus of the center $(-g, -f)$,let $x = -g$ and $y = -f$,so $g = -x$ and $f = -y$.Substituting these into the equation: $a^{2} + b^{2} + 2(-x)a + 2(-y)b + 4 = 0$.This simplifies to $a^{2} + b^{2} - 2ax - 2by + 4 = 0$,or $2ax + 2by - (a^{2} + b^{2} + 4) = 0$.

If a circle passes through the point $(a, b)$ and intersects the circle $x^{2} + y^{2} = 4$ orthogonally,then the locus of its center is:

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