Question

Find the angle of intersection of the following curve:
$x^2+y^2=2x$ and $y^2=x$

Answer 1

$x^2+y^2=2x$, $y^2=x$

solving them, $x^2+x=2x$

$x^2=x$

$x=0,1$

$\Rightarrow y=0$, at $x=0$

$\Rightarrow atx=1,y=\pm 1$

$\therefore (0,0),(1,1),(1,-1)$ are point of intersection

Now, for curve (1) differentiating is

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

$x+y{y}^{\prime }=1$

$y^{\prime }=\frac{1-x}{y}$

$\left(m_1\right)$

fro curve (2)

$2y{y}^{\prime }=1$

$y^{\prime }=\frac{1}{2y}$

$\left(m_2\right)$

for $(1,1)$ $m_1=0m_2=\frac{1}{2}$

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$=\left|\frac{0-\frac{1}{2}}{1}\right|=\frac{1}{2}$

$\theta ={\tan }^{-1}\frac{1}{2}$

for ($1,-1)$ $m_1=0m_2=-\frac{1}{2}$

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

$\theta ={\tan }^{-1}\frac{1}{2}$