The locus of the centre of a circle which cut | vedclass

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(C) Let the required circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since this circle cuts the circle $x^2 + y^2 + 4x - 6y + 9 = 0$ orthogonally,we use the

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$8x - 12y + 5 = 0$

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(C) Let the required circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since this circle cuts the circle $x^2 + y^2 + 4x - 6y + 9 = 0$ orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$:$2g(2) + 2f(-3) = c + 9 \Rightarrow 4g - 6f = c + 9$ .....$(i)$Since it also cuts $x^2 + y^2 - 4x + 6y + 4 = 0$ orthogonally:$2g(-2) + 2f(3) = c + 4 \Rightarrow -4g + 6f = c + 4$ .....$(ii)$Adding $(i)$ and $(ii)$ gives $0 = 2c + 13$,so $c = -\frac{13}{2}$.Subtracting $(ii)$ from $(i)$ gives $8g - 12f = 5$.The centre of the circle is $(-g, -f)$. Let the locus be $(x, y)$,so $x = -g$ and $y = -f$,which means $g = -x$ and $f = -y$.Substituting these into the equation $8g - 12f = 5$ gives $8(-x) - 12(-y) = 5$,which simplifies to $-8x + 12y = 5$ or $8x - 12y + 5 = 0$.

The locus of the centre of a circle which cuts the circles $x^2 + y^2 + 4x - 6y + 9 = 0$ and $x^2 + y^2 - 4x + 6y + 4 = 0$ orthogonally is

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