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

Toolbox:
• If $y=f(x),$ then $\bigg(\large\frac{dy}{dx} \bigg)_p$ = slope of the tangent to $y=f(x)$ at point $p$
Step 1
$y^2=4x\: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \:\: \: \: \:\: \: \: (1)$
$x^2+y^2-6x+1=0\: \: (2)$
Let us find the points of intersection by solving the two equations :
substituting equation (1) in equation (2) we get,
$x^2+4x-6x+1=0$
$\Rightarrow x^2-2x+1=0$
$\Rightarrow (x-1)^2=0$
$\Rightarrow x=1\: and \: y= \pm 2$
Hence the points of intersection are $(1,2)\: and \: (1, -2)$
Step 2
Consider the curve $c_1$
$y^2=4x$
differentiate w.r.t $x$
$2y\large\frac{dy}{dx}=4$
$\Rightarrow \large\frac{dy}{dx}=\large\frac{4}{2y}=\large\frac{2}{y}$
$\bigg( \large\frac{dy}{dx}\bigg)_{(1,2)}=\large\frac{2}{2}=1$
This is the slope of the curve $c_1$ at (1,2) is 1
Step 3
Consider the curve $c_2$
$x^2+y^2-6x+1=0$
differentiate w.r.t $x$
$2x+2y\large\frac{dy}{dx}-6=0$
$\Rightarrow 2y\large\frac{dy}{dx}=6-2x$
$\therefore \large\frac{dy}{dx}=\large\frac{6-2x}{2y}$
$\Rightarrow \large\frac{dy}{dx}=\large\frac{3-x}{2}$
$\bigg( \large\frac{dy}{dx}\bigg)_{(1,2)}= \large\frac{3-1}{2}$
$=1$
That is the slope of the curve $c_2$ is 1
Since the slope of the curves at the point (1,2) is the same, they touch each other.