A circle of constant radius `a` passes through the origin `O` and cuts the axes of coordinates at points `P` and `Q` . Then the equation of the locus of the foot of perpendicular from `O` to `P Q` is `(x^2+y^2)(1/(x^2)+1/(y^2))=4a^2` `(x^2+y^2)^2(1/(x^2)+1/(y^2))=a^2` `(x^2+y^2)^2(1/(x^2)+1/(y^2))=4a^2` `(x^2+y^2)(1/(x^2)+1/(y^2))=a^2`

Options

A. `(x^(2)+y^(2))((1)/(x^(2))+(1)/(y^(2)))=4a^(2)`
B. `(x^(2)+y^(2))^(2)((1)/(x^(2))+(1)/(y^(2)))=a^(2)`
C. `(x^(2)+y^(2))^(2)((1)/(x^(2))+(1)/(y^(2)))=4a^(2)`
D. `(x^(2)+y^(2))((1)/(x^(2))+(1)/(y^(2)))=a^(2)`

Answer

3

Explanation

The equation of line PQ is
`y-k=-(h)/(k)(x-h)`
or `hx+ky=h^(2)+k^(2)`
`:. Q=((h^(2)+k^(2))/(h),0)`
and `P-=(0,(h^(2)+k^(2))/(k))`
Also, `2a=sqrt(x_(1)^(2)+y_(1)^(2))`
or `x_(1)^(2)+y_(1)^(2)=4a^(2)`

Eliminating `x_(1)` and `y_(1)`, we have
`(x^(2)+y^(2))((1)/(x^(2))+(1)/(y^(2)))=4a^(2)`