The product of the slopes of the common tangents drawn to the circles $x^2+y^2+2 x-2 y-2=0$ and $x^2+y^2-2 x+$

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Aicharya · Accepted Answer

Let $S_1: x^2+y^2+2 x-2 y-2=0$ $\begin{aligned} & C_1 \equiv(-1,1) \text { and } r_1=\sqrt{1+1+2}=2 \\ & S_2: x^2+y^2-2 x+2 y+1=0 \\ & C_2 \equiv(1,-1) \text { and } r_2=\sqrt{1+1-1}=1\end{aligned}$

$\because \frac{\mathrm{C}_1 \mathrm{P}}{\mathrm{C}_2 \mathrm{P}}=\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{2}{1} \Rightarrow \mathrm{C}_1 \mathrm{P}=2 \mathrm{C}_2 \mathrm{P}$ $\begin{aligned} & \text { Now, } \mathrm{C}_1 \mathrm{C}_2+\mathrm{C}_2 \mathrm{P}=\mathrm{C}_1 \mathrm{P} \Rightarrow \mathrm{C}_1 \mathrm{C}_2+\mathrm{C}_2 \mathrm{P}=2 \mathrm{C}_2 \mathrm{P} \\ & \Rightarrow \mathrm{C}_1 \mathrm{C}_2=\mathrm{C}_2 \mathrm{P}\end{aligned}$ So, $\mathrm{C}_2$ is mid point of $\mathrm{C}_1$ and $\mathrm{P}$. $\begin{aligned} & \Rightarrow \frac{-1+x_1}{2}=1 \text { and } \frac{y_1+1}{2}=-1 \\ & \Rightarrow x_1=3, y_1=-3\end{aligned}$ $\therefore \quad \mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)=(3,-3)$ Equation of line passing through $\mathrm{P}$ is : $(y+3)=m(x-3) \Rightarrow m x-y-3 m-3=0$ ...(i) $\because \quad r_2=$ Perpendicular distance from $(1,-1)$ to line (i) $\Rightarrow 1=\left|\frac{\mathrm{m}+1-3 \mathrm{~m}-3}{\sqrt{\mathrm{m}^2+1}}\right| \Rightarrow 3 \mathrm{~m}^2+8 \mathrm{~m}+3=0$ $\begin{aligned} & \Rightarrow \mathrm{m}=\frac{-8 \pm \sqrt{28}}{6} \\ & \therefore \mathrm{m}_1=\frac{-8+\sqrt{28}}{6} \text { and } \mathrm{m}_2=\frac{-8-\sqrt{28}}{6}\end{aligned}$ $\Rightarrow m_1 m_2=\frac{64-28}{36}=1$

The product of the slopes of the common tangents drawn to the circles $x^2+y^2+2 x-2 y-2=0$ and $x^2+y^2-2 x+$

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