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