If the circles $x^2+y^2=9$ and $x^2+y^2+2 \alpha x+2 y+1=0$ touch each other internally, then the value of $\alpha^3$ is

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Aicharya · Accepted Answer

$\begin{aligned} & x^2+y^2=9 \ & \mathrm{C}_1=(0,0), \mathrm{r}_1=3 \ & x^2+y^2+2 \alpha x+2 y+1=0 \ & \mathrm{C}_2=(-\alpha,-1), \ & \mathrm{r}_2=\sqrt{\alpha^2+1-1}=\alpha \end{aligned}$ Since the given circles touch each other internally, $\begin{aligned} & \mathrm{C}_1 \mathrm{C}_2=\left|\mathrm{r}_1-\mathrm{r}_2ight| \ & \Rightarrow \sqrt{\mathrm{a}^2+1}=|3-\alpha| \ & \Rightarrow \alpha^2+1=9+\alpha^2-6 \alpha \ & \Rightarrow 6 \alpha=8 \ & \Rightarrow \alpha=\frac{4}{3} \ & \Rightarrow \alpha^3=\frac{64}{27} \end{aligned}$

If the circles $x^2+y^2=9$ and $x^2+y^2+2 \alpha x+2 y+1=0$ touch each other internally, then the value of…

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