Let C be a circle that passes through the origin. Show that we can find real numbers s and t such that C is the graph of

Question

r = 2s*cos(theta + t)

I think that I could start by writing it like R^2=(x-h)^2+(y-k)^2 where R is the radius of the circle, but I don't what to do with that info.

Answer 1

consider the circle
r = 2cosθ
that is (x-1)^2+y^2 = 1

clearly
r = 2scosθ

is just a scaled up version of that.

(x - s)^2 + y^2 = s^2

r = 2scos(θ+t)

is just a rotated version, turned through angle t.

Since t can be anything from 0 to 2pi, the center of our circle can be anywhere on the circle (x-s)^2+y^2 = s^2

or something.

Answer 2

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