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Q.
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
NTA AbhyasNTA Abhyas 2022
Solution:
Equation of tangent to circle $x^{2}+y^{2}=1$ is $y=mx\pm\sqrt{1 + m^{2}}$ . This also touches the circle $\left(x - 2\right)^{2}+y^{2}=4$ .
$\therefore \left|\frac{2 m \pm \sqrt{1 + m^{2}}}{\sqrt{1 + m^{2}}}\right|=2\Rightarrow m=\pm\sqrt{\frac{1}{3}}$ .
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 $x=-2$
$\Rightarrow h=-2$ .