JEE Advance - Mathematics (2001 - No. 24)
The equation of the common tangent touching the circle $${\left( {x - 3} \right)^2} + {y^2} = 9$$ and the parabola $${y^2} = 4x$$ above the $$x$$-axis is
$$\sqrt {3y} = 3x + 1$$
$$\sqrt {3y} = - \left( {x + 3} \right)$$
$$\sqrt {3y} = x + 3$$
$$\sqrt {3y} = - \left( {3x + 1} \right)$$