The locus of the centre of circles passing th | vedclass

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(C) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since it passes through $(a, b)$,we have $a^2+b^2+2ga+2fb+c=0$,which implies $c = -a^2-b^

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$(a-1)x+(b+2)y=\frac{a^2+b^2+4}{2}$

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(C) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since it passes through $(a, b)$,we have $a^2+b^2+2ga+2fb+c=0$,which implies $c = -a^2-b^2-2ga-2fb$. Since the circle cuts $x^2+y^2-2x+4y-4=0$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1+c_2$ gives: $2g(-1) + 2f(2) = c - 4$. Substituting $c$: $-2g + 4f = -a^2-b^2-2ga-2fb - 4$. Rearranging terms: $g(2a-2) + f(2b+4) = -a^2-b^2-4$. Let the centre be $(x, y) = (-g, -f)$,so $g = -x$ and $f = -y$. Substituting these: $-x(2a-2) - y(2b+4) = -a^2-b^2-4$. $x(2a-2) + y(2b+4) = a^2+b^2+4$. Dividing by $2$: $(a-1)x + (b+2)y = \frac{a^2+b^2+4}{2}$.

The locus of the centre of circles passing through $(a, b)$ and cutting the circle $x^2+y^2-2x+4y-4=0$ orthogonally is

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