Q93 / 180 · KCET 2008 · Mathematics
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 $\alpha$ is equal to
- Correct answerA$\pm \frac{4}{3}$
- B1
- C$\frac{4}{3}$
- D$-\frac{4}{3}$
Solution
Centres and radii of the given circles $x^{2}+y^{2}=9$ and $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \alpha \mathrm{x}+2 \mathrm{y}+1=0$ is $\mathrm{C}_{1}(0,0), \mathrm{r}_{1}=3$ and $C_{2}(-\alpha, 1)$ and $r_{2}=\sqrt{\alpha^{2}+1-1}=|\alpha|$
Since, two circles touch internally,
$$
\begin{array}{lrl}
\therefore & \mathrm{C}_{1} \mathrm{C}_{2} & =\mathrm{r}_{1}-\mathrm{r}_{2} \\
\Rightarrow & \sqrt{\alpha^{2}+1^{2}} & =3-|\alpha|
\end{array}
$$
$\Rightarrow \quad \alpha^{2}+1=9+\alpha^{2}-6|\alpha|$
$\Rightarrow \quad 6|\alpha|=8 \Rightarrow|\alpha|=\frac{4}{3}$
$\Rightarrow \quad \alpha=\pm \frac{4}{3}$
Hence, the correct answer is option: 'A'.