JEE Main 2019 — Circle Question with Solution

Since, circle passing through origin intersect the coordinate axes at $A\& B$, hence $AB$ must be diameter and $AB=2R$.

Now, let foot of the perpendicular from origin upon $AB$ be $P\left(h, k\right)$.

Slope of line $OP=\frac{k-0}{h-0}=\frac{k}{h}$

Since, line $AB\perp OP\Rightarrow$ slope of $AB=-\frac{h}{k}$

Thus, equation of line  $AB$ is $y-k=\frac{-h}{k}\left(x-h\right)$

For co-ordinates of $A$, put $y=0\Rightarrow 0-k=\frac{-h}{k}\left(x-h\right)\Rightarrow x=\frac{h^2+k^2}{h}\Rightarrow A \left(\frac{h^2+k^2}{h}, 0\right)$.

For co-ordinates of $B$, put $x=0\Rightarrow y-k=\frac{-h}{k}\left(0-h\right)\Rightarrow y=\frac{h^2+k^2}{k}\Rightarrow B \left(0,\frac{h^2+k^2}{k}\right)$.

Now, given $AB=2R$

Applying distance formula,

$\Rightarrow \sqrt{{\left(\frac{h^2+k^2}{h}-0\right)}^2+{\left(0-\frac{h^2+k^2}{k}\right)}^2}=2R$

$\Rightarrow {\left(h^2+k^2\right)}^2 \left(\frac{1}{h^2}+\frac{1}{k^2}\right)=4R^2\Rightarrow {\left(h^2+k^2\right)}^3=4R^2{h}^2{k}^2$

Hence, locus is ${\left(x^2+y^2\right)}^3=4R^2{x}^2{y}^2$