IIT JEE Advanced 2008 Paper 1

We have
C1 : y2 = 4x (1)
C2 : x2+y2 - 6x + 1 = 0 (2)
Substituting y2 = 4x in Eq. (2), we get
x2 + 4x - 6x + 1 = 0
x2 - 2x + 1 = 0
x = 1
That is, the points of intersection of the circle and the parabola are (1, 2) and (1, –2).

Clearly, C1 and C2 touch each other exactly at two points.