Question

If the circles $x^2+y^2-2x=0$ and $x^2+y^2-2\lambda y=4$ have only one common tangent then $\lambda$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is C $0$

Let $S_1=x^2+y^2-2x=0$ and $S_2=x^2+y^2-2\lambda y=4$
$\Rightarrow C_1=(1,0),r_1=1,C_2=(0,\lambda ),r_2=\sqrt{{\lambda }^2+4}$
Now for above circle to have only one common tangent,
$r_2-r_2=C_1{C}_2\Rightarrow \sqrt{{\lambda }^2+4}-1=\sqrt{1+{\lambda }^2}$
$\Rightarrow {\lambda }^2+4+1-2\sqrt{{\lambda }^2+4}=1+{\lambda }^2$
$\Rightarrow {\lambda }^2+4=4\Rightarrow \lambda =0$