Question

find the center and radius of a circle whose equation in polar coordinates is r=3 cos(theta)+10 sin(theta)

Answer 1

There may be a more direct way, but here's what I'd do:
we have the coordinate conversion y = r sin(theta) and x = r cos(theta)
Hence sin(theta) = y/r and cos(theta) = x/r, and we substitute these in to get
\[r = {3x \over r} + {10y \over r}\]\[r^2 = 3x + 10y\]Now r^2 = x^2+y^2, so we have
\[x^2 + y^2 = 3x + 10y\]\[x^2 - 3x + y^2 - 10y = 0\]Completing the square on both gives\[x^2 -3x + {9 \over 4} + y^2 -10y + 25 = {109 \over 4}\]And factoring gives us what we need:\[(x-{3 \over 2})^2 + (y-5)^2 = ({\sqrt{109} \over 2})^2\]Thus the center is (3/2, 5) and the radius is sqrt(109)/2

Answer 2

THANK YOU! I was missing the second step