Question

From a point $\left(h,0\right)$ common tangents are drawn to circles $x^2+y^2=1$ and ${\left(x-2\right)}^2+y^2=4$ , value of $h$ is

• A. $2$
• B. $-2$
• C. $-\frac{2}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

Equation of tangent to circle $x^2+y^2=1$ is $y=mx\pm \sqrt{1+m^2}$ . This also touches the circle ${\left(x-2\right)}^2+y^2=4$ .
$\therefore \left|\frac{2m\pm \sqrt{1+m^2}}{\sqrt{1+m^2}}\right|=2\Rightarrow m=\pm \sqrt{\frac{1}{3}}$ .
Therefore common tangents are
$y=\frac{x}{\sqrt{3}}+\frac{2}{\sqrt{3}}$ and $y=\frac{-x}{\sqrt{3}}-\frac{2}{\sqrt{3}}$
On putting $y=0$ , from both equations we get $x=-2$
$\Rightarrow h=-2$ .