The tagents at any point P of the circle `x^(2)+y^(2)=16` meets the tangents at a fixed point A at T. Point is joined to B, the other end of the diameter, through, A.
The locus of the intersection of AO and BT is a conic whosee eccentricity is

Options

A. `1//2`
B. `1//sqrt(2)`
C. `1//3`
D. `1//sqrt(3)`

Explanation

Tangent at `P(4 cos theta, 4 sin theta) "to x^(2)+y^(2)=16` is `x cos thetay+sin theta=4 " "(1)`
The equation of AP is `y=(sin theta)/(cos theta-1)(x-4)" "(2)`

From (1), the coordinates of the point T are given by
`(4,(4(1-costheta))/(sin theta))`
The equation of BT is `y=(1-cos theta)/(2 sin theta)(x+4)" "(3)`
Let (h,k) be the point of intersection of the lines (2), and (3). Then we have
`k^(2)=-(1)/(2)(h^(2)-16)`
or `(h^(2))/(16)+(y^(2))/(8)=1`
Therefore, the locus of (h,k) is `(x^(2))/(16)+(y^(2))/(8)=1`
Which is an ellipse with eccehntrically `e=1//sqrt(2)`
Sum of focal distance of any points is 2a=8
Considering the circle `x^(2)+y^(2)=a^(2)`, we find the that the eccentricity of the ellipse is `1sqrt(2)` which is contant and does not change by changing the radius of the circle