Question

A circle of constant radius 'a' passes through origin 'O' and cuts the axes of co-ordinates in points P and Q, then the equation of the locus of the foot of perpendicular from O to PQ is

• A. $(x^2+y^2)(\frac{1}{x^2}+\frac{1}{y^2})=4a^2$
• B. $(x^2+y^2{)}^2(\frac{1}{x^2}+\frac{1}{y^2})=a^2$
• C. $(x^2+y^2{)}^2(\frac{1}{x^2}+\frac{1}{y^2})=4a^2$
• D. $(x^2+y^2)(\frac{1}{x^2}+\frac{1}{y^2})=a^2$

Answer 1

Answer: (C)

$(x^2+y^2{)}^2(\frac{1}{x^2}+\frac{1}{y^2})=4a^2$

Equation of line PQ is y - k $=-\frac{h}{k}(x-h)$
or $hx+ky=h^2+k^2$
$\Rightarrow$ Points $Q(\frac{h^2+k^2}{h},0)$ and $P(0,\frac{h^2+k^2}{k})$
Also, $2a=\sqrt{x_1^2+y_1^2}$
$\Rightarrow x_1^2+y_1^2=4a^2$
Eliminating $x_1$ and $y_1$ we have
$(x^2+y^2{)}^2(\frac{1}{x^2}+\frac{1}{y^2})=4a^2$