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(C) Let the required equation of the circle be $x^2+y^2+2gx+2fy=0$ (since it passes through the origin,$c=0$).Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$

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$3x^2+3y^2-8x+29y=0$

Vedclass · Answer

(C) Let the required equation of the circle be $x^2+y^2+2gx+2fy=0$ (since it passes through the origin,$c=0$).Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.For the first circle $x^2+y^2-6x+8=0$,$g_1=-3, f_1=0, c_1=8$. Thus,$2g(-3)+2f(0)=0+8$ $\Rightarrow -6g=8$ $\Rightarrow g=-\frac{4}{3}$.For the second circle $x^2+y^2-2x-2y-7=0$,$g_2=-1, f_2=-1, c_2=-7$. Thus,$2g(-1)+2f(-1)=0-7 \Rightarrow -2g-2f=-7$.Substituting $g=-\frac{4}{3}$,we get $-2(-\frac{4}{3})-2f=-7$ $\Rightarrow \frac{8}{3}+7=2f$ $\Rightarrow 2f=\frac{29}{3}$.Substituting $g$ and $f$ into the general equation: $x^2+y^2+2(-\frac{4}{3})x+2(\frac{29}{6})y=0 \Rightarrow x^2+y^2-\frac{8}{3}x+\frac{29}{3}y=0$.Multiplying by $3$,we get $3x^2+3y^2-8x+29y=0$.

The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-6x+8=0$ and $x^2+y^2-2x-2y-7=0$ is

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