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

Answer: 3

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

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