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)

Centres and radii of the given circles $x^2+y^2=9$
and $x^2+y^2+2ax+2y+1=0$ 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,
$\therefore C_1{C}_2=r_1-r_2$
$\Rightarrow ,\sqrt{{\alpha }^2+1^2}=3-|\alpha |$
$\Rightarrow {\alpha }^2+1=9+{\alpha }^2-6|\alpha |$
$\Rightarrow 6|\alpha |=8$
$\Rightarrow |\alpha |=\frac{4}{3}$
$\Rightarrow \alpha =\pm \frac{4}{3}$