Question

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

• A. $2 \sqrt{3} y = 12 \; x + 1$
• B. $2 \sqrt{3} y = - x - 12 $
• C. $\sqrt{3} y = x + 3$
• D. $\sqrt{3} y = 3x + 1$

Answer 1

Answer: (C)

Let equation of tangent to the parabola $y^2 = 4x$ is
$y=mx + \frac{1}{m} \Rightarrow m^{2}x -ym +1= 0$ is tangent to $ x^{2}+y^{2} -6x=0$
$ \Rightarrow \frac{\left|3m^{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$