Equation of an Oﬀ-Center Circle This is a standard example that comes up a lot. Circles are easy to describe, unless the origin is on the rim of the circle. We’ll calculate the equation in polar coordinates of a circle with center (a, 0) and radius (2a, 0). You should expect to repeat this calculation a few times in this class and then memorize it for multivariable calculus, where you’ll need it often. (a,0) Figure 1: Oﬀ center circle through (0, 0). In rectangular coordinates, the equation of this circle is: (x − a)2 + y 2 = a2 . We could plug in x = r cos θ, y = sin θ to convert to polar coordinates, but there’s a faster way. We start by expanding and simplifying: (x − a)2 + y 2 = a2 x2 − 2ax + a2 + y 2 = a2 x2 − 2ax + y 2 = (x2 + y 2 ) − 2ax = r2 − 2ar cos θ = 0 0 0 2 = 2ar cos θ =⇒ r = 2a cos θ r (or r = 0). We used the facts that x2 + y 2 = r2 and x = r cos θ to conclude that there were two values of r that satisfy this equation; r = 2a cos θ and r = 0. These are the equations describing r in terms of θ that describe this circle in polar coordinates. In order to use the equation r = 2a cos θ, we need to ﬁgure out the appro priate range of values for θ. By looking at the graph we see that − π2 < θ < π2 . Our ﬁnal equation is: r=0 or r = 2a cos θ, 1 − π π <θ< . 2 2 y r θ (a,0) x Figure 2: Oﬀ center circle in polar coordinates. To check our work, let’s ﬁnd some points on this curve: • At θ = 0, r = 2a and so x = 2a and y = 0. √ • At θ = π4 , r = 2a cos π4 = a 2. Hence x = a and y = a. 2