Question

Tangent and normal to a curve, at any point $P$ meet $x$ and $y$ axes at $A , B$ and $C , D$ respectively also centre of the circle through $O , C , P$ and $B$ lies on the line $x + y =0$. If the curve passes through $(1,0)$, and equation of the curve is given by $\left(x^2+y^2\right)=\mu e^{k \tan ^{-1} \frac{y}{x}}$, then find the value of $(\mu+k)$.

Answer 1

$T: Y-y=\frac{d y}{d x}(X-x)$
$N: Y-y=\frac{-d x}{d y}(X-x) $
$B \equiv\left(0, y-x \frac{d y}{d x}\right)$
$C \equiv\left(x+y \frac{d y}{d x}, 0\right)$
$BC$ is diameter centre lies on $y + x =0$
$x+y \frac{d y}{d x}=-\left(y-x \frac{d y}{d x}\right) $
$x+y=(x-y) \frac{d y}{d x}$

$\frac{d y}{d x}=\frac{x+y}{x-y} \text {. Put } y=V x $
$V + x \frac{ dV }{ dx }=\frac{1+ V }{1- V }$
$X \frac{ dV }{ dx }=\frac{1+ V - V + V ^2}{1- V }$
$\frac{1-V}{1+V^2} d V=\frac{d x}{x} $
$\tan ^{-1} V-\frac{1}{2} \ln \left(1+V^2\right)=\ln x+\ln c $
$2 \tan ^{-1} \frac{y}{x}=\ln \left(x^2+y^2\right) \cdot C^2$
$C ^2\left( x ^2+ y ^2\right)= e ^{ y / x } $
$\text { Passes }(1,0) $
$C ^2= e ^0 \Rightarrow C ^2=1$
$\left(x^2+y^2\right)=e^{2 \tan ^{-1} y / x}$
$\therefore \mu=1, k =2 $
$\mu+k=3 . $