Expert: Sherry Wallin - 1/28/2011

Question
P(x,y) is an arbitary point on the circle x squared + y squared=36 express the distance from P to the point (0,8) a a function of the x-coordinate of P

Answer
Sarah Smith~

First, notice that (0,8) is NOT on the circle described by
x^2+y^2 = 36. This is a circle centered at (0,0) six units from
the origin in every direction. You want an equation in x that describes the distance from any x on the circle to the point (0,8).

Solve for y from x^2 + y^2 = 36 giving you y = +-sqrt(36-x^2)
and using the distance formula with (0,8) and (+-sqrt(36-x^2) we have:

d((x,y),(0,8)) = sqrt[(x-0)^2 + (+-sqrt(36-x^2)-8)^2]
= sqrt(x^2 +36-x^2 -16*sqrt(36-x^2)+64
= sqrt(100-16*sqrt(36-x^2))

thus

f(x) = sqrt(100-16*sqrt(36-x^2)) which is a function of x!

Math Prof