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

We have, ${y^2} = 4x$ and ${x^2} + {y^2} - 6x + 1 = 0$

Since, both the curves touch each other at (1,2) i.e., curves are passing through (1,2) .

Therefore, $2y \cdot \frac{{dy}}{{dx}} = 4$
and $2x + 2y\frac{{dy}}{{dx}} = 6$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{4}{{2y}}$

and $\frac{{dy}}{{dx}} = \frac{{6 - 2x}}{{2y}}$

$\Rightarrow$ $\quad {\left( {\frac{{dy}}{{dx}}} \right)_{(1,2)}} = \frac{4}{4} = 1$

and ${\left( {\frac{{dy}}{{dx}}} \right)_{(1,2)}} = \frac{{6 - 2 \cdot 1}}{{2 \cdot 2}} = \frac{4}{4} = 1$

$\Rightarrow$ ${m_1} = 1$ and ${m_2} = 1$

Thus, we see that slope of both the curves are equal to each other i.e., ${m_1} = {m_2} = 1$ at the point (1,2).

Hence, both the curves touch each other.