Question

The set of value of $\alpha$ for which the circles given by $x^2+y^2={\alpha }^2$ and $x^2+y^2-2\sqrt{2}x-2\sqrt{2}y-\alpha =0$ have exactly three common tangents is

• A. $-2,4$
• B. $0,5$
• C. $2$
• D. $3$

Answer 1

The correct option is B $0,5$

$x^2+y^2={\alpha }^2$
$C_1=(0,0)$ & $r_1=\alpha$
$x^2+y^2-2\sqrt{2}x-2\sqrt{2}y-\alpha =0$
$C_2=(\sqrt{2},\sqrt{2})$ & $r_2=\sqrt{4+\alpha }$
Since, circles have exactly three common tangents
Therefore, circles touch each other and thus $C_1{C}_2=r_1+r_2$
$\sqrt{2+2}=\alpha +\sqrt{4+\alpha }$
$\Rightarrow {(2-\alpha )}^2=4+\alpha$
$\Rightarrow {\alpha }^2-4\alpha =\alpha$
$\Rightarrow \alpha =0,5$