If the circles $x^2+y^2-4 x+2 f y+1=0$ and $x^2+y^2+$ $2 \mathrm{gx}-4 \mathrm{y}-1=0$ cut orthogonally, then $\mathrm{r}_1^2+\mathrm{r}_2^2-8=$

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Aicharya · Accepted Answer

Let $S_1: x^2+y^2-4 x+2 f y+1=0$ ...(i) $S_2: x^2+y^2+2 g x-4 y-1=0$ ...(ii) $\begin{aligned} & \mathrm{g}_1=2, \mathrm{f}_1=\mathrm{f}, \mathrm{c}_1=1 \ & \mathrm{~g}_2=\mathrm{g}, \mathrm{f}_2=2, \mathrm{c}_2=-1\end{aligned}$ $\because$ Circle (i) \& (ii) are orthogonal : $2\left(g_1 g_2+f_1 f_2ight)=c_1+c_2$ and $r_1^2+r_2^2=d^2$ where $d$ is distance between the centers $\begin{aligned} & 	ext { Now, } 2\left(\mathrm{f}_1 \mathrm{f}_2+\mathrm{g}_1 \mathrm{~g}_2ight)=\mathrm{c}_1+\mathrm{c}_2 \ & \Rightarrow 2(2 \mathrm{f}+2 \mathrm{~g})=1-1 \Rightarrow \mathrm{f}=-\mathrm{g} \ & 	ext { Also, } \mathrm{r}_1{ }^2+\mathrm{r}_2{ }^2=\mathrm{d}^2\end{aligned}$ $\begin{aligned} & \Rightarrow r_1^2+r_2^2=(g-2)^2+(g+2)^2 \ & \Rightarrow r_1^2+r_2^2=2 g^2+8 \Rightarrow r_1^2+r_2^2-8=2 g^2\end{aligned}$

If the circles $x^2+y^2-4 x+2 f y+1=0$ and $x^2+y^2+$ $2 \mathrm{gx}-4 \mathrm{y}-1=0$ cut orthogonally,…

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