# Plotting polar equations of circles not centered at (0, 0)

Good afternoon guys!

I'm fairly new to polar coordinates and polar equations, so bear with me please. I understand the equation of a circle with radius $a$ centered at the polar coordinate $(r_0, \phi)$ is as follows:

$$r = r_0cos(\theta-\phi)+\sqrt{a^2-r_0^2sin^2(\theta-\phi)}$$

Where $(r, \theta)$ represents any arbitrary point on the circle. I understand how to derive this equation from Cartesian coordinates as well, and I can recognize how the equation works with relative ease. My problem comes when I try to graph the circle. How would one go about doing it?

Take the following example:

Let $r_0=5$, $a=2$, and $\phi=tan(3/4)$, meaning we are representing a circle with radius $2$ centered at the Cartesian coordinate $(4, 3)$. If we sub in the values, we get:

$$r = 5cos(\theta-tan(3/4))+\sqrt{4-25sin^2(\theta-tan(3/4))}$$

Now if the circle was centered at $(0, 0)$, to plot such a thing we could simply just start at $\theta=0$ and work our way up to $\theta=2\pi$, subbing in the values as we go and solving for the new radius (which would always be $a$). But since this circle isn't centered at $(0, 0)$, it means that certain lines with angle $\theta$ will never intersect the circle. In this case, if we subbed in $\theta=0$, then $\sqrt{4-25sin^2(\theta-tan(3/4))}$ would evaluate to a complex number, since the line at angle $\theta=0$ does not intersect the circle described by the equation above. Obviously this isn't something we want.

So where do you start plotting such an equation? How do you know which angles to use and which angles to avoid in order to get the points on the circle?

If you are asking how to plot the circle, well, you know the centre and radius so you just draw it! You don't need to know which angles to use and which to avoid.

On the other hand, perhaps I have missed the point and maybe you do want to know the angles for some other reason. If so you can find them as follows.

First case: if the origin is inside the circle then any $\theta$ will be valid.

Second case: the origin is outside the circle. Draw the circle with centre $C$, draw the line from the origin $O$ to $C$, and a tangent from the origin to the circle, meeting the circle at $T$. Then $CTO$ is a right angle, so the angle $\alpha$ between $OC$ and $OT$ is given by $$\sin\alpha=\frac{CT}{OC}\ .$$ Using the notation in your question, this is $$\sin\alpha=\frac{a}{r_0}$$ and the possible values are $$\phi-\alpha\le\theta\le\phi+\alpha\ .$$

Final case: if the origin actually lies on the circle, then from a diagram you can see that we must have $$\phi-\frac{\pi}{2}\le\theta\le\phi+\frac{\pi}{2}\ .$$

• Haha, I really wish it was as easy as just drawing it! I'm working on plotting polar equations of curves using a computer, so I was looking for a mathematical definition, which you presented quite well. This definitely clears up my confusion, so thank you very much! Apr 3 '14 at 1:00

That's it. You want to avoid non-real numbers, so avoid $\theta : 4−25\sin^2(\theta−\tan(3/4))<0$

$$\therefore \text{Find } \theta : \left|\sin(\theta-\tan\left(\frac 34\right))\right| \leq \frac 2 5$$

$$\implies \theta : \left|\theta - \tan\left(\frac 34\right)\right| \leq \arcsin \left(\frac 2 5\right) \color{gray}{+ 2 n \pi, \quad \exists n \in \mathbb{Z}}$$

$$\implies \theta : \tan\left(\frac 34\right)-\arcsin\left(\frac 25\right)\leq \theta \color{gray}{- 2 n \pi} \leq \tan\left(\frac 34\right)+\arcsin \left(\frac 2 5\right)$$