Prove that the curves $y^{2}=4x$ and $x^{2}+y^{2}-6x+1=0$ touch each other at the point $(1,2)$.

(N/A) Given curves are $C_1: y^{2}=4x$ and $C_2: x^{2}+y^{2}-6x+1=0$.
First,we verify if the point $(1,2)$ lies on both curves:
For $C_1: (2)^{2} = 4(1) \implies 4 = 4$ (True).
For $C_2: (1)^{2} + (2)^{2} - 6(1) + 1 = 1 + 4 - 6 + 1 = 0$ (True).
Now,we find the slopes of the tangents at $(1,2)$ by differentiating both equations with respect to $x$.
For $C_1: 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$.
At $(1,2)$,$m_1 = \frac{2}{2} = 1$.
For $C_2: 2x + 2y \frac{dy}{dx} - 6 = 0 \implies \frac{dy}{dx} = \frac{6-2x}{2y} = \frac{3-x}{y}$.
At $(1,2)$,$m_2 = \frac{3-1}{2} = \frac{2}{2} = 1$.
Since the slopes of the tangents to both curves at the point $(1,2)$ are equal $(m_1 = m_2 = 1)$,the curves touch each other at the given point.