Question:
Equation of a common tangent to the circle, $x^{2}+y^{2}-6 x=0$ and the parabola, $y^{2}=4 x$, is:
Correct Option: , 3
Solution:
Let equation of tangent to the parabola $y^{2}=4 x$ is
$y=m x+\frac{1}{m}$
$\Rightarrow m^{2} x-y m+1=0$ is tangent to $x^{2}+y^{2}-6 x=0$
$\Rightarrow \frac{\left|3 m^{2}+1\right|}{\sqrt{m^{4}+m^{2}}}=3$
$m=\pm \frac{1}{\sqrt{3}}$
$\Rightarrow$ tangent are $x+\sqrt{3} y+3=0$
and $x-\sqrt{3} y+3=0$