JEE Main 12 Jan 2019 Evening Question 24
Question: 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]
Options:
A) $ {{(x^{2}+y^{2})}^{3}}=4R^{2}x^{2}y^{2} $
B) $ {{(x^{2}+y^{2})}^{2}}=4Rx^{2}y^{2} $
C) $ (x^{2}+y^{2})(x+y)=R^{2}xy $
D) $ {{(x^{2}+y^{2})}^{2}}=4R^{2}x^{2}y^{2} $
Show Answer
Answer:
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 $ ( \frac{h^{2}+k^{2}}{h},0 ) $ and $ ( 0,\frac{h^{2}+k^{2}}{k} ) $ respectively.
Now, $ AB=2R\Rightarrow {{(AB)}^{2}}=4R^{2} $
$ \Rightarrow $ $ \frac{{{(h^{2}+k^{2})}^{2}}}{h^{2}}+\frac{{{(h^{2}+k^{2})}^{2}}}{k^{2}}=4R^{2} $
$ \Rightarrow $ $ {{(h^{2}+k^{2})}^{3}}=4R^{2}h^{2}k^{2} $
Thus, locus of P is $ {{(x^{2}+y^{2})}^{3}}=4R^{2}x^{2}y^{2}. $
