Equation of a common tangent to the circle, x2+y2−6x=0 and the parabola,y2=4x, is

The given equation of circle is

$x^{2} + y^{2} - 6 x = 0 \Rightarrow (x - 3)^{2} + (y - 0)^{2} = (3)^{2}$ $\dots (1)$

Here, centre is (3, 0) and radius = 3

Let equation of tangent to the parabola $y^{2} = 4 x$ is $y = m x + \frac{1}{m}$ $\dots (2)$

$\Rightarrow m^{2} x - m y + 1 = 0$ which is also tangent to (i) so, distance from point (3, 0) to the tangent = Radius

$∴ |\frac{3 m^{2} + 1}{\sqrt{m^{4} + m^{2}}}| = 3 \Rightarrow m = \pm \frac{1}{\sqrt{3}}$

$∴$ Required equation of tangents are

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

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