Equation of a common tangent to the circle,

Question:

Equation of a common tangent to the circle, $x^{2}+y^{2}-6 x=0$ and the parabola, $y^{2}=4 x$, is:

  1. $2 \sqrt{3} y=12 x+1$

  2. $2 \sqrt{3} \mathrm{y}=-\mathrm{x}-12$

  3. $\sqrt{3} y=x+3$

  4. $\sqrt{3} y=3 x+1$


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 $x-\sqrt{3} y+3=0$

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