Mathematics - Circle Question with Solution | TestHub

Let $(h,k)$ be the point on the curve $y={\tan }^{-1}x$.

Image of $(h,k)$ in $y=x$ is $(k,h)$ which is the centre of a circle of radius $\frac{\pi }{2\sqrt{2}}$

Given $PM=\frac{\pi }{2\sqrt{2}}$ (shortest distance)

And $CM=\frac{\pi }{2\sqrt{2}}$ (radius of circle)

Now,

$$CP=\sqrt{(h-k{)}^2+(k-h{)}^2}=\frac{\pi }{2}$$

$$\begin{array}{rlr}& \Rightarrow \sqrt{2}|h-k|=\frac{\pi }{\sqrt{2}}\Rightarrow |h-k|=\frac{\pi }{2}& \end{array}$$

$$\Rightarrow h-k=\pm \frac{\pi }{2}\Rightarrow k=h\pm \frac{\pi }{2}$$

Since, $(h,k)$ lies on $y={\tan }^{-1}x$

$$\Rightarrow k=h-\frac{\pi }{2}$$

Now, $h=\frac{x}{c}={\tan }^{-1}h$

Since,

$$(1<h<\pi \Rightarrow \frac{-\pi }{2}<h-\frac{\pi }{2}<\frac{\pi }{2}$$

$$h=\mathrm{an}\left(h-\frac{\pi }{2}\right)=-\coth \Rightarrow -h=\coth$$