Prove that the curves

Given curve equations are: y2 = 4x …. (1) and x2 + y2 – 6x + 1 = 0 ….. (i)

Now, differentiating (i) w.r.t. x, we get

2y.(dy/dx) = 4 ⇒ dy/dx = 2/y

Slope of tangent at (1, 2), m1 = 2/2 = 1

Differentiating (ii) w.r.t. x, we get

2x + 2y.(dy/dx) – 6 = 0

2y. dy/dx = 6 – 2x ⇒ dy/dx = (6 – 2x)/ 2y

Hence, the slope of the tangent at the same point (1, 2)

⇒ m2 = (6 – 2 x 1)/ (2 x 2) = 4/4 = 1

It’s seen that m1 = m2 = 1 at the point (1, 2).

Therefore, the given circles touch each other at the same point (1, 2).