JEE Main 2019 — Mathematics Coordinate Geometry

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

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

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

Since, line $AB\perp OP\Rightarrow$AB⊥OP⇒ slope of $AB=-\frac{h}{k}$AB=−kh​

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

For co-ordinates of $A$A, put $y=0\Rightarrow 0-k=\frac{-h}{k}(x-h)\Rightarrow x=\frac{{h}^{2}+{k}^{2}}{h}\Rightarrow A (\frac{{h}^{2}+{k}^{2}}{h},0)$y=0⇒0−k=k−h​(x−h)⇒x=hh2+k2​⇒A(hh2+k2​,0).

For co-ordinates of $B$B, put $x=0\Rightarrow y-k=\frac{-h}{k}(0-h)\Rightarrow y=\frac{{h}^{2}+{k}^{2}}{k}\Rightarrow B (0,\frac{{h}^{2}+{k}^{2}}{k})$x=0⇒y−k=k−h​(0−h)⇒y=kh2+k2​⇒B(0,kh2+k2​).

Now, given $AB=2R$AB=2R

Applying distance formula,

$$\Rightarrow \sqrt{{(\frac{{h}^{2}+{k}^{2}}{h}-0)}^{2}+{(0-\frac{{h}^{2}+{k}^{2}}{k})}^{2}}=2R$$⇒(hh2+k2​−0)2+(0−kh2+k2​)2​=2R

$$\Rightarrow {({h}^{2}+{k}^{2})}^{2} (\frac{1}{{h}^{2}}+\frac{1}{{k}^{2}})=4{R}^{2}\Rightarrow {({h}^{2}+{k}^{2})}^{3}=4{R}^{2}{h}^{2}{k}^{2}$$⇒(h2+k2)2(h21​+k21​)=4R2⇒(h2+k2)3=4R2h2k2

Hence, locus is ${({x}^{2}+{y}^{2})}^{3}=4{R}^{2}{x}^{2}{y}^{2}$(x2+y2)3=4R2x2y2