Question

If the circles $x ^2 + y ^2 =9$ and $x^2 + y^ 2 + 2\alpha x + 2y +1 = 0$ touch each other internally, then $\alpha$ =

• A. $\pm \frac {4}{3}$
• B. $1$
• C. $\frac {4}{3}$
• D. $- \frac {4}{3}$

Answer 1

Answer: (A)

$\pm \frac {4}{3}$

Answer 2

The correct option is(A): $\pm \frac {4}{3}$.

Centers and radii of the given circles $x^{2}+y^{2}=9$
and $x^{2}+y^{2}+2 a x+2 y+1=0$ is $C_{1}(0,0), r_{1}=3$
and $C_{2}(-\alpha, 1)$ and $r_{2}=\sqrt{\alpha^{2}+1-1}=|\alpha|$
Since, two circles touch internally,
$\therefore C_{1} C_{2}=r_{1}-r_{2}$
$\Rightarrow \,\,\,,\sqrt{\alpha^{2}+1^{2}}=3-|\alpha|$
$\Rightarrow \,\,\,\alpha^{2}+1=9+\alpha^{2}-6|\alpha|$
$\Rightarrow \,\,\,\,6|\alpha|=8$
$\Rightarrow|\alpha|=\frac{4}{3}$
$\Rightarrow \,\,\,\alpha=\pm \frac{4}{3}$