Question

A normal is drawn at a point $P(x,y)$ on a curve. If it meets the $x-$axis and the $y-$axis such that $(x-\text{intercept}{)}^{-1}+(y-\text{intercept}{)}^{-1}=1,$ then the radius of the director circle of the curve passing through $(3,3)$ is

Answer 1

Let $O$ be the origin and $A,B$ be the points where the curve intersects at $x$ and $y$ axes respectively.
Equation of the normal at $P(x,y)$ is
$Y-y=\frac{-1}{\frac{dy}{dx}}(X-x)$
$\Rightarrow (X-x)+(Y-y)\frac{dy}{dx}=0$
Thus, $x-\text{intercept}=OA=x+y\frac{dy}{dx}$
and $y-\text{intercept}=OB=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}$
Given that $\frac{1}{OA}+\frac{1}{OB}=1$
$\Rightarrow 1+\frac{dy}{dx}=x+y\frac{dy}{dx}$
$\Rightarrow (y-1)dy+(x-1)dx=0$
Integrating both sides, we get
$(x-1{)}^2+(y-1{)}^2=C$
Since curve is passing through the $(3,3),$
$\therefore 4+4=C\Rightarrow C=8$
$\therefore (x-1{)}^2+(y-1{)}^2=8$
which is equation of a circle of radius $2\sqrt{2}.$
Radius of director circle $=\sqrt{2}\times 2\sqrt{2}=4$