If `AB` is the intercept of the tangent to the circle `x^2 +y^2=r^2` between the coordinate axes, the locus of the vertex `P` of the rectangle `OAPB` is

Options

A. `x^(2)+y^(2)=r^(2)`
B. `(1)/(x^(2))+(1)/(y^(2))=(1)/(r^(2))`
C. `(1)/(x^(2))+(1)/(y^(2))=r^(2)`
D. none of these

Answer

B

Explanation

The equation of any tangent to `x^(2)+y^(2)=r^(2)` is `x cos theta + y sin theta = r`.

This meets the coordinates axes at A (r sec `theta`, 0) and B (0, r cosec `theta`).
`:. H =r sec theta and k = r cosec theta`
`rArr cos theta = (r)/(h) and sin theta = (r)/(k)`
`rArr cos^(2)theta+sin^(2)theta = (r^(2))/(h^(2))+(r^(2))/(k^(2))rArr (1)/(h^(2))+(1)/(k^(2))=(1)/(r^(2))`