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

Options

A. `sqrt(3) y = 3x + 1`
B. `2 sqrt(3) y = 12x + 1`
C. `sqrt(3) y = x + 3`
D. `2 sqrt(3) y = - x - 12`

Answer

C

Explanation

We know that, equation of tangent to parabola `y^(2) = 4ax`
`y = mx + (a)/(m)`
`:.` Equation of tangent to the parabola `y^(2) = 4x` is
`y = mx + (1)/(m)` `( :' a = 1)`
`implies m^(2) x - my + 1 = 0`
Now, let line (i) is also a tangent to the circle.
Equation of circle `x^(2) + y^(2) - 6x = 0`
Clearly, centre of given circle is (3, 0) and radius = 3
[`:'` for the circle `x^(2) + y^(2) + 2 gx + fy + c = 0`, centre = (-g, -f) and radius `= sqrt(g^(2) + f^(2) -c]`
`:.` The perpendicular distance of (3,0) from the line (i) is 3.
[`:'` Radius is perpendicular to the tangent of cirlce]
`implies (|m^(2).3 - m.0 + 1|)/(sqrt((m^(2))^(2) + (-m)^(2))) = 3`
The length of perpendicular from a point `(x_(1), y_(1))` to the line `ax + by + c = 0` is `|(ax_(1) + by_(1) + c)/(sqrt(a^(2) + b^(2)))|`
`implies (3m^(2) + 1)/(sqrt(m^(4) + m^(2))) = 3`
`implies 9m^(4) + 6m^(2) + 1 = 9 (m^(4) + m^(2))`
`implies m ~~ oo` or `m = +- (1)/(sqrt(3))`
`[:' underset(m to oo)("lim") (3m^(2) + 1)/(sqrt(m^(4) + m^(2))) = underset(m to oo)("lim") (3 + (1)/(m^(2)))/(sqrt(1 + (1)/(m^(2))) = 3]`
`:. ` Equation of common tangents are x = 0
`y = (x)/(sqrt(3)) + sqrt(3)` and `y = (-x)/(sqrt(3)) - sqrt(3)`