find the exact coordinates of the point on a circle of radius 5 at an angle of 5pi/3. thanks for any help! (:

Answer 1

I will assume you want your circle to centre (0,0), then its equation is
x^2 + y^2 = 25
2x + 2y dy/dx = 0
dy/dx = -x/y <---- the slope of the tangent
you want the tangent angle to be 5π/3 or 300°
tan 5π/3 = -√3

then -x/y = -√3
x = √3 y
subbing back into my circle equation
3y^2 + y^2 = 25
4y^2 = 25
y = ± 5/2
if y = 5/2, then y = 5√3/2 -------> (5√3/2 , 5/2)
if y = -5/2, then y = -5√3/2 -----> ( -5√3/2 , -5/2)

https://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+y%5E2+%3D+25+%2C+y+-+5%2F2+%3D+-%E2%88%9A3%28x+-+5%E2%88%9A3%2F2%29%2C+y+%2B+5%2F2+%3D+-%E2%88%9A3%28x+%2B+5%E2%88%9A3%2F2%29

Answer 2

if your circle has center at (0,0) then you have
x = r cosθ
y = r sinθ
Now just plug in your numbers