ExamBro
ExamBro
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

  1. A \(\frac{27}{64}\)
  2. B \(\frac{125}{27}\)
  3. C \(\frac{27}{125}\)
  4. D \(\frac{64}{27}\)
Verified Solution

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,
\(\begin{aligned}
& \mathrm{C}_1 \mathrm{C}_2=\left|\mathrm{r}_1-\mathrm{r}_2\right| \\
& \Rightarrow \sqrt{\mathrm{a}^2+1}=|3-\alpha| \\
& \Rightarrow \alpha^2+1=9+\alpha^2-6 \alpha \\
& \Rightarrow 6 \alpha=8 \\
& \Rightarrow \alpha=\frac{4}{3} \\
& \Rightarrow \alpha^3=\frac{64}{27}
\end{aligned}\)