Let `S` be the circle in the `x y` -plane defined by the equation `x^2+y^2=4.` (For Ques. No 15 and 16) Let `P` be a point on the circle `S` with both coordinates being positive. Let the tangent to `S` at `P` intersect the coordinate axes at the points `M` and `N` . Then, the mid-point of the line segment `M N` must lie on the curve `(x+y)^2=3x y` (b) `x^(2//3)+y^(2//3)=2^(4//3)` (c) `x^2+y^2=2x y` (d) `x^2+y^2=x^2y^2`

Options

A. `(x+y)^(2) = 3xy`
B. `x^(2//3)+y^(2//3) =2^(4//3)`
C. `x^(2)+y^(2)=2xy`
D. `x^(2)+y^(2)=x^(2)y^(2)`

Answer

4

Explanation

Let the coordinates of P be `(2 cos theta , 2 sin theta )`
Equation of tangent to circle at P is `x cos x + y sin theta =2`
`:. M((2)/(cos theta ,0)),N(0,(2)/(cos theta))`
Let the mid -point of MN be `(x,y)`
`:. x= (1)/(cos theta ) ` and `y = (1)/(sin theta)`
Squaring and adding , we get
`(1)/(x^(2))+(1)/(y^(2))=1 `
or `x^(2)+y^(2)=x^(2)y^(2)`, this is the required locus