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 sum of focal distance of any point on the curce is

Options

A. 12
B. 16
C. 20
D. 8

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