Question

Answer the following :

Find the equation of tangent to the circle x² + y² = 64 at the point `"P"((2pi)/3)`

Answer 1

Given equation of circle is

x² + y² = 64

Comparing this equation with x² + y² = r², we get

r = 8

The equation of a tangent to the circle

x² + y² = r² at P(θ) is

x cosθ + y sinθ = r

∴ the equation of the tangent at `"P"((2pi)/3)` is

`xcos  (2pi)/3 + y sin  (2pi)/3` = 8

∴ `x((-1)/2) + y(sqrt(3)/2)` = 8

∴ `-x + sqrt(3)y` = 16

∴ `x - sqrt(3)y + 16` = 0.