MHT CET · Maths · Circle
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
- A \(\frac{27}{64}\)
- B \(\frac{125}{27}\)
- C \(\frac{27}{125}\)
- D \(\frac{64}{27}\)
Answer & Solution
Correct Answer
(D) \(\frac{64}{27}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& x^2+y^2=9 \\
& \mathrm{C}_1=(0,0), \mathrm{r}_1=3 \\
& x^2+y^2+2 \alpha x+2 y+1=0 \\
& \mathrm{C}_2=(-\alpha,-1), \\
& \mathrm{r}_2=\sqrt{\alpha^2+1-1}=\alpha
\end{aligned}\)
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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