Q63 / 180 · COMEDK 2019 · Mathematics
If the circle $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+20 x+2 y+1=0$ touch each other internally, then $\alpha=$
- A$\pm \frac{4}{3}$
- B1
- Correct answerC$\frac{4}{3}$
- D$\frac{-4}{3}$
Solution
Let $\quad S_{1}: x^{2}+y^{2}=9$
and $\quad S_{2}: x^{2}+y^{2}+2 \alpha x+2 y+1=0$
Now, centre and radius of $S_{1}$ are
Centre $=C_{1}(0,0)$
Radius $=3$
Centre and radius of $S_{2}$ are
$$
\begin{aligned}
&\text { Centre }=C_{2}(-\alpha,-1) \\
&\text { Radius }=\sqrt{(-\alpha)^{2}+(-1)^{2}-1}=\alpha
\end{aligned}
$$
Since, $S_{1}$ and $S_{2}$ touch each other internally, then
$$
\begin{array}{ll}
& C_{1} C_{2}=\left|r_{2}-r_{1}\right| \\
\Rightarrow & \sqrt{(-\alpha-0)^{2}+(-1-0)^{2}}=|\alpha-3| \\
\Rightarrow & \sqrt{\alpha^{2}+1}=|\alpha-3| \\
\Rightarrow & \alpha^{2}+1=(\alpha-3)^{2} \\
\Rightarrow & \alpha^{2}+1=\alpha^{2}-6 \alpha+9 \\
\Rightarrow & 6 \alpha=8 \Rightarrow \alpha=\frac{4}{3}
\end{array}
$$
Hence, the correct answer is option: 'C'.