Question

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

Options

• `2sqrt(3)y = 12x + 1`

• `sqrt(3)y = x + 3`

• `2sqrt(3)y = - x - 12`

• `sqrt(3)y = 3x + 1`

Answer 1

`sqrt(3)y = x + 3`

Explanation:

Since, the equation of tangent to parabola `y^2 = 4x` is `y = mx + 1/m`  .....(i)

The line (1) is also the tangent to circle `x^2 + y^2 - 6x` = 0

Then centre of circle = (3, 0)

Radius of circle = 3

The radius of the circle is equal to the perpendicular distance from the centre to the tangent.

∴ `(|3m + 1/m|)/sqrt(1 + m^2)` = 3

⇒ `(3m + 1/m)^2 = 9(1 + m^2)`

⇒ `m = +- 1/sqrt(3)`

Then, from equation (1): `y = +- 1/sqrt(3) x +- sqrt(3)`

Hence, `sqrt(3)y = x + 3` is one of the required common tangent.