The locus of the point of intersection of the tangents to the circle `x^2+ y^2 = a^2` at points whose parametric angles differ by `pi/3`.

Options

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

Answer

C

Explanation

Let the parametric angles of two points on the circle `x^(2)+y^(2)=a^(2)` be `theta` and `pi//3+theta`. Then, the two points are `P(a cos theta, a sin theta)` and `Q (a cos (pi//33+theta), a sin (pi//3+theta))`.
The equations of the tangents at P and Q are
`x cos theta + y sin theta 0 = a " " (i) `
and , `x cos (pi //3 + theta) + y sin (pi//3 + 0)=a`, respectively
Now,
`x cos (pi//3 + theta)+y sin (pi//3+theta)=a`
`rArr (1)/(2)(x cos theta + y sin theta)-(sqrt(3))/(2)(x sin theta-y cos theta)=a`
`rArr (a)/(2)-(sqrt(3))/(2)(x sin theta-y cos theta)=a " " `[Using (i)]
`rArr x sin 0- y cos theta = -(a)/(sqrt(3)) " " ...(ii)`
The locus to the point of intersection of the two tangents is obtained by eliminating `theta` between (i) and (ii).
From (i) and (ii), we get
`(x cos theta + y sin theta)^(2)+(x sin theta - y cos theta)^(2)=a^(2)+((a)/(sqrt(3)))^(2)`
`rArr x^(2)+y^(2)=4 a^(2)//3rArr(x^(2)+y^(2))=4a^(2)`