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A very long time ago in algebra/trig class we did polar equation of a circle where

$r = 2a\cos\theta + 2b\sin\theta$

Now I forgot how to derive this. So I tried using the standard form of a circle.

$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$

$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$

$$(a^2\cos^2 \theta + a^2 - 2a^2\cos\theta) + (b^2\sin^2 \theta + b^2 - 2b^2\sin\theta) = a^2 + b^2$$

$$a^2\cos^2 \theta + b^2\sin^2 \theta - 2a^2 \cos\theta - 2b^2 \sin\theta = 0$$

Now I am stuck, I think I was supposed to complete the square or something. Could someone finish my thought?

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Hem, $x=r\cos\theta$, not $a\cos\theta$. – Yves Daoust Jul 17 '15 at 7:59
up vote 6 down vote accepted

I think your substitutions from the first line to the second aren't quite right. It looks like you used $x=a\cos\theta$ and $y=b\sin\theta$, but you probably wanted $$\begin{align}x&=r\cos\theta\\y&=r\sin\theta.\end{align}$$ Using those, $$\begin{align} (x-a)^2 + (y - b)^2 &= a^2 + b^2 \\ (r\cos\theta-a)^2+(r\sin\theta-b)^2&=a^2+b^2 \\ r^2\cos^2\theta+a^2-2ar\cos\theta+r^2\sin^2\theta+b^2-2br\sin\theta&=a^2+b^2 \\ r^2(\sin^2\theta+\cos^2\theta)+(a^2+b^2)-2r(a\cos\theta+b\sin\theta)&=a^2+b^2 \\ r^2-2r(a\cos\theta+b\sin\theta)&=0 \\ r^2=2r(a\cos\theta+b\sin\theta) \\ r=0\;\;\text{ or }\;\;r=2a\cos\theta+2b\sin\theta \end{align}$$ and since $r=0$ describes only the pole, which is also contained in the graph of $r=2a\cos\theta+2b\sin\theta$, $$r=2a\cos\theta+2b\sin\theta$$ is sufficient to describe the circle.

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I find it easier to go from polar to rectangular.

$r=2a\cos\theta+2b\sin\theta$; $r^2=2ar\cos\theta+2br\sin\theta$; $x^2+y^2=2ax+2by$; $x^2-2ax+y^2-2by=0$; $(x-a)^2+(y-b)^2=a^2+b^2$; voila, circle of radius $\sqrt{a^2+b^2}$ centered at $(a,b)$.

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... and there's the completing-the-square, which is absent when going in the other direction. +1 – Isaac Jun 6 '12 at 4:59

A 'backwards' answer:

The general polar equation of a circle of radius $\rho$ centered at $(r_0,\theta_0)$ is $$r^2-2 r r_0 \cos(\theta-\theta_0) + r_0^2 = \rho^2.$$ When $r_0 = \rho$, this reduces to (ignoring the $r=0$ solution) $$r = 2(r_0 \cos \theta_0) \cos \theta + 2(r_0 \sin \theta_0) \sin \theta,$$ choosing $(r_0,\theta_0)$ such that $a = r_0 \cos \theta_0$, $b = r_0 \sin \theta_0$ gives the required form.

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It should be noted that this equation of the circle is simply the Law of Cosines with appropriate variables. – eh9 Jun 7 '15 at 14:29

I agree with the answer given by Gerry Myerson. The only detail left out is, is to complete the square on the terms: $x^2-2ax = (x-a)^2 - a^2$ and the same for: $y^2-2by = (y-b)^2 - b^2$ adding these 2 equations and equal to zero then gives the equation in Cartesian form as shown above by Gerry Myerson. I am just adding in the details for those that may not realize.

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