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(B) Let the equation of the circle be $S \equiv x^2+y^2+2gx+2fy=0$ (since it passes through the origin $(0,0)$).For the circle $S_1 \equiv x^2+y^2+6x-

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

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(B) Let the equation of the circle be $S \equiv x^2+y^2+2gx+2fy=0$ (since it passes through the origin $(0,0)$).For the circle $S_1 \equiv x^2+y^2+6x-15=0$,we have $2g_1=6, 2f_1=0, c_1=-15$. The condition for orthogonality is $2gg_1+2ff_1=c+c_1$,which gives $2g(3)+2f(0)=0-15$ $\Rightarrow 6g=-15$ $\Rightarrow g=-\frac{5}{2}$.For the circle $S_2 \equiv x^2+y^2-8y-10=0$,we have $2g_2=0, 2f_2=-8, c_2=-10$. The condition for orthogonality is $2gg_2+2ff_2=c+c_2$,which gives $2g(0)+2f(-4)=0-10$ $\Rightarrow -8f=-10$ $\Rightarrow f=\frac{5}{4}$.Substituting $g$ and $f$ into the equation $S$,we get $x^2+y^2+2(-\frac{5}{2})x+2(\frac{5}{4})y=0$.Multiplying by $2$,we get $2x^2+2y^2-10x+5y=0$.

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

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