# Find the angle to intersection of the following curves:

Question:

Find the angle to intersection of the following curves:

$x^{2}+y^{2}=2 x$ and $y^{2}=x$

Solution:

Given:

Curves $x^{2}+y^{2}=2 x \ldots(1)$

$\& y^{2}=x \ldots(2)$

Solving $(1) \&(2)$, we get

Substituting $y^{2}=x$ in $x^{2}+y^{2}=2 x$

$\Rightarrow x^{2}+x=2 x$

$\Rightarrow x^{2}-x=0$

$\Rightarrow x(x-1)=0$

$\Rightarrow x=0$ or $(x-1)=0$

$\Rightarrow x=0$ or $x=1$

Substituting $x=0$ or $x=1$ in $y^{2}=x$, we get,

when $x=0$

$\Rightarrow y^{2}=0$

$\Rightarrow y=0$

when $x=1$

$\Rightarrow y^{2}=1$

$\Rightarrow y=1$

The point of intersection of two curves are $(0,0) \&(1,1)$

Now, Differentiating curves (1) \& (2) w.r.t $x$, we get

$\Rightarrow x^{2}+y^{2}=2 x$

$\Rightarrow 2 x+2 y \cdot \frac{d y}{d x}=2$

$\Rightarrow x+y \cdot \frac{d y}{d x}=1$

$\Rightarrow y \cdot \frac{d y}{d x}=1-x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-\mathrm{x}}{\mathrm{y}} \ldots(3)$

$\Rightarrow y^{2}=x$

$\Rightarrow 2 y \cdot \frac{d y}{d x}=1$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2 y} \ldots(4)$

At $(1,1)$ in equation $(3)$, we get

$\Rightarrow \frac{d y}{d x}=\frac{1-x}{y}$

$\Rightarrow \frac{d y}{d x}=\frac{1-1}{1}$

$\Rightarrow \mathrm{m}_{1}=0$

At $(1,1)$ in equation $(4)$, we get

$\Rightarrow \frac{d y}{d x}=\frac{1}{2 y}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2 \times 1}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}$

$\Rightarrow m_{2}=\frac{1}{2}$

when $m_{1}=0 \& m_{2}=\frac{1}{2}$

$\Rightarrow \tan \theta=\left|\frac{0-\frac{1}{2}}{1+0 \times \frac{1}{2}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{-1}{2}}{1+0}\right|$

$\Rightarrow \tan \theta=\left|\frac{-1}{2}\right|$

$\Rightarrow \tan \theta=\frac{1}{2}$

$\Rightarrow \theta=\tan ^{-1}\left(\frac{1}{2}\right)$

$\Rightarrow \theta \cong 26.56$