The equation of the common tangent touching the circle `(x-3)^2+y^2=9` and the parabola `y^2=4x` above the x-axis is

Options

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

Answer

C

Explanation

Any tengent to `y^(2) = 4x` is of the form `y = mx + (1)/(m)`
(`:' a =1`) and this touches the circle `(x - 3)^(2) + y^(2) = 9`
If `|(m(33) + (1)/(m) - 0)/(sqrt(m^(2) + 1))| = 3`
[`:'` centre of the circle is (3,0) and radisu is 3]
`implies (3m^(2) + 1)/(m) = +- 3 sqrt(m^(2) + 1)`
`implies 3m^(2) + 1 = +- 3m sqrt(m^(2) + 1)`
`implies 9m^(4) + 1 + 6m^(2) = 9m^(2) (m^(2) + 1)`
`implies 9m^(4) + 1 + 6m^(2) = 9m^(4) + 9m^(2)`
`implies 3m^(2) = 1`
`implies m = +- (1)/(sqrt(3))`
If the tangent touches the parabola and circle above the X-axis, then slope m should be positive.
`:. m = (1)/(sqrt(3))` and the equation is `y = (1)/(sqrt(3)) x + sqrt(3)`
or `sqrt(3) y = x + 3`