Question

If the number of common tangents to the pair of circles $x^{2}+y^{2}-2 x+4 y-4=0$ and $x^{2}+y^{2}+4 x-4 y+\alpha=0$ is 4, then the least integral value of $\alpha$ is

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (B)

We have,
$x^{2}+y^{2}-2 x+4 y-4=0$
and $x^{2}+y^{2}+4 x-4 y+\alpha=0$
$c_{1}=(1,-2), r_{1}=\sqrt{1+4+4}=3$
$c_{2}=(-2,2), r_{2}=\sqrt{4+4-\alpha}=\sqrt{8-\alpha}$
Circle have 4 common tangents
$\therefore \,c_{1} c_{2}>\,r_{1}+r_{2}$
$\Rightarrow \, \sqrt{(1+2)^{2}+(-2-2)^{2}}>3+\sqrt{8-\alpha}$
$\Rightarrow \, \sqrt{9+16}>3+\sqrt{8-\alpha}$
$\Rightarrow \, 2>\, \sqrt{8-\alpha}$
$\Rightarrow \, 8-\alpha<\,4$
$\Rightarrow \, \alpha>\,4$
$\therefore $ Least integral value of $\alpha=5$