question_answer

If a circle of radius R passes through the origin and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from 0 on AB is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

A) \[{{({{x}^{2}}+{{y}^{2}})}^{3}}=4{{R}^{2}}{{x}^{2}}{{y}^{2}}\]                

B) \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=4R{{x}^{2}}{{y}^{2}}\]

C) \[({{x}^{2}}+{{y}^{2}})(x+y)={{R}^{2}}xy\]                    

D) \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=4{{R}^{2}}{{x}^{2}}{{y}^{2}}\]

Correct Answer: A

Solution :

Let P(h, k) be the foot of the perpendicular from O to AB. \[\therefore \] Slope of \[AB=\frac{-h}{k}\] \[\therefore \]Equation of AB is          \[hx+ky={{h}^{2}}+{{k}^{2}}\] \[\Rightarrow \]\[\frac{x}{\frac{{{h}^{2}}+{{k}^{2}}}{h}}+\frac{y}{\frac{{{h}^{2}}+{{k}^{2}}}{k}}=1\] So, coordinates of A and B are \[\left( \frac{{{h}^{2}}+{{k}^{2}}}{h},0 \right)\]and \[\left( 0,\frac{{{h}^{2}}+{{k}^{2}}}{k} \right)\]respectively. Now,\[AB=2R\Rightarrow {{(AB)}^{2}}=4{{R}^{2}}\] \[\Rightarrow \]\[\frac{{{({{h}^{2}}+{{k}^{2}})}^{2}}}{{{h}^{2}}}+\frac{{{({{h}^{2}}+{{k}^{2}})}^{2}}}{{{k}^{2}}}=4{{R}^{2}}\] \[\Rightarrow \]\[{{({{h}^{2}}+{{k}^{2}})}^{3}}=4{{R}^{2}}{{h}^{2}}{{k}^{2}}\] Thus, locus of P is \[{{({{x}^{2}}+{{y}^{2}})}^{3}}=4{{R}^{2}}{{x}^{2}}{{y}^{2}}.\]