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### Answers (1)

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

Hint - The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$

m1=slope of first curve.   m2=slope of second curve.
Given –
\begin{aligned} &x^{2}+y^{2}=2 x \ldots \ldots(1) \\ &y^{2}=x \ldots \ldots \text { (2) } \end{aligned}

Considering second curve

\begin{aligned} &y^{2}=x \end{aligned}

Substituting this in eq (1)
$x^{2}+x=2x\\ x^{2}+x-2x=0\\ x^{2}-x=0\\ x\left (x-1 \right )=0\\ x = 0\;\;\; or\;\;\; x-1 =0\\ x =0\;\;\; or\;\;\;x = 1\\$

Now putting the value of x = 0 & 1 in  \begin{aligned} &y^{2}=x \end{aligned}
When x = 0

$y^{2}=0\\ y=0$
When x = 1

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

Point of intersection are (0, 0) and (1, 1)

Since curves are  $x^{2}+y^{2}=2 x \;\;\&\;\; y^{2}=x$

Differentiating above with respect to x

As we know,  $\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$

\begin{aligned} &=x^{2}+y^{2}=2 x \\ &=2 x+2 y \frac{d y}{d x}=2 \\ &=2 y \frac{d y}{d x}=2-2 x \end{aligned}

\begin{aligned} &=\frac{d y}{d x}=\frac{2-2 x}{2 y}=\frac{2(1-x)}{2 y} \\ &m_{1}=\frac{d y}{d x}=\frac{1-x}{y} \end{aligned}

Now consider  \begin{aligned} &y^{2}=x \end{aligned}
\begin{aligned} &2 y \frac{d y}{d x}=1 \\ &=m_{2}=\frac{d y}{d x}=\frac{1}{2 y} \end{aligned}

When (0, 0)
\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{1-0}{0}=\infty \\ &=m_{2}=\frac{d y}{d x}=\frac{1}{2(0)}=\infty \end{aligned}

When (1, 1)

\begin{aligned} &=m_{1}=\frac{1-1}{1}=0 \\ &=m_{2}=\frac{1}{2(1)}=\frac{1}{2} \end{aligned}

Angle of intersection of two curves is given by$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
When m1 is  $\frac{4}{3}$  and  m2  is  $\frac{1}{3}$
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{tan} \theta=\left|\frac{0-\frac{1}{2}}{1+0\left(\frac{1}{2}\right)}\right| \\ &\operatorname{tan} \theta=\left|\frac{\frac{-1}{2}}{1}\right| \\ &\operatorname{tan} \theta=\left(\frac{1}{2}\right) \end{aligned}

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