Question

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

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (B)

We have,
$x^2+y^2-2x+4y-4=0$
and $x^2+y^2+4x-4y+\alpha =0$
$c_1=(1,-2),r_1=\sqrt{1+4+4}=3$
$c_2=(-2,2),r_2=\sqrt{4+4-\alpha }=\sqrt{8-\alpha }$
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 $\alpha =5$