Correct Answer - C
Equation of given circles is `x^(2)+y^(2)=1`, then equation of tangent at the point `(costheta, sintheta)` on the given circle is
`xcostheta+y sintheta=1" "...(i)`
[`therefore` Equation of tangent at the point `P(costheta, sintheta)` to the circle `x^(2) +y^(2)=r^(2) is x cos theta+ y sintheta=r`]
Now, the point of intersection with coordinates axes are `P(sectheta, 0) and Q(0,cosec theta)`
`therefore` Mid-point of line joining points P and Q is
`M((sectheta)/(2),(cosectheta)/(2))=(h,k)` (let)
`So, costheta=(1)/(2h)and sintheta=(1)/(2k)`
`therefore sin^(2)theta+cos^(2)theta=1`
`therefore (1)/(4h^(2))+(1)/(4k^(2))=1rArr(1)/(h^(2))+(1)/(k^(2))=4`
Now, locus of mid-point M is
`(1)/(x^(2))+(1)/(y^(2))=4`
`rArr x^(2)+y^(2)-4x^(2)y^(2)=0`