The equation of the circle passing through (0 | vedclass

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(A) Let the equation of the circle passing through $(0,0)$ be $x^2+y^2+2gx+2fy=0$ ...$(i)$.The given circles are $x^2+y^2+6x-15=0$ ...$(ii)$ and $x^2+

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$2(x^2+y^2)-10x+5y=0$

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(A) Let the equation of the circle passing through $(0,0)$ be $x^2+y^2+2gx+2fy=0$ ...$(i)$.The given circles are $x^2+y^2+6x-15=0$ ...$(ii)$ and $x^2+y^2-8y-10=0$ ...$(iii)$.Since circle $(i)$ cuts circle $(ii)$ orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$:$2(g)(3) + 2(f)(0) = 0 + (-15)$ $\Rightarrow 6g = -15$ $\Rightarrow g = -\frac{5}{2}$.Since circle $(i)$ cuts circle $(iii)$ orthogonally:$2(g)(0) + 2(f)(-4) = 0 + (-10)$ $\Rightarrow -8f = -10$ $\Rightarrow f = \frac{5}{4}$.Substituting $g$ and $f$ into equation $(i)$:$x^2+y^2+2(-\frac{5}{2})x+2(\frac{5}{4})y = 0$.$x^2+y^2-5x+\frac{5}{2}y = 0$.Multiplying by $2$,we get $2(x^2+y^2)-10x+5y=0$.

The equation of the circle passing through $(0,0)$ and cutting orthogonally the circles $x^2+y^2+6x-15=0$ and $x^2+y^2-8y-10=0$ is

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