How to write the equation of an off-center circle in terms of r (parameterization)

Question

I've looked around online for a while and I can't seem to find a question that properly captures what I'm talking about (or answers that really help me).

For a homework problem, I have to find the area bounded by $x^{2} + y^{2} = 4$ and $(x-1)^{2} + (y-1)^2 = 1$ using double integrals in polar form. I can easily parameterize the first equation as just $r = 2$ using simple algebra.

However, for the second equation, I get stuck quickly. I can first expand the equation to get $x^{2} -2x + 1 + y^{2} -2y + 1 = 1$. I can simplify to get $x^2 + y^2 + 1 = 2(x + y)$. From here I can substitute the polar equivalents of x and y, to get

$r^2 + 1 = 2\big(rcos(\theta) + rsin(\theta)\big)$

From here I have no idea what to do. The math to get here seems simple enough, yet it's somehow wrong.

Once I have the second equation in terms of $r$, I'd imagine I could set the two equations equal to each other, and find at what angles they are equal (giving me the range of $\theta$ over which I need to integrate), then I'd have the second integral for $r$ to be over the bounds of the two equations. But then, I don't know what function I'd be integrating over that specific range.

Thanks for any help. I really am at a loss as to where to turn

Answer 1

In fact this polar equation is in two parts ($r_1$ and $r_2$: see equations (2) and (3) below) which is not surprising when you consider figure 1.

Write your equation under the form:

$$r^2 - 2r\underbrace{(\cos(\theta) + \sin(\theta))}_{f(\theta)}+1=0 \tag{0}$$

and consider it as a quadratic equation, or even simpler as factorisable under the form:

$$[r-f(\theta)]^2+\underbrace{1-f(\theta)^2}_{-2 \sin(\theta)\cos(\theta)}=0$$

Consequently

$$[r-f(\theta)]^2=2 \sin(\theta)\cos(\theta)=\sin(2\theta)$$

which means:

$$r-f(\theta)=\pm \sqrt{\sin(2\theta)}\tag{1}$$

under the condition that $\theta$ belongs to the first quadrat:

$$0\le \theta \le \pi/2$$

in order that $\sin(2\theta) \ge 0$ (needed for taking the square root in (1)).

The previous expressions gives rise to 2 polar equations:

$$r_1=\cos(\theta) + \sin(\theta)-\sin(2\theta) \ \ \ \ \text{(blue arc below)}\tag{2}$$

$$r_2=\cos(\theta) + \sin(\theta)+\sin(2\theta) \ \ \ \ \text{(red arc below)}\tag{3}$$

Fig. 1: the two circles.

Important remark: you need only the blue arc of the second circle.

Another remark: If you want to consider (0) as a quadratic equation, you will find the same $r_1$ and $r_2$ as its roots using discriminant:

$$\Delta=4((\cos(\theta) + \sin(\theta))^2-1)=4(2\cos(\theta)\sin(\theta))=4 \sin(2\theta) $$

Answer 2

Introduce two different polar coordinates (let's call them $s$ and $\phi$, just so that we don't confuse them with $r$ and $\theta$), where:

• $s$ is the length of a vector pointing from the center of the circle to some point in space, and
• $\phi$ is the angle that vector makes with the positive $x$-axis.

Then the conversion between these new coordinates and the usual rectangular coordinates is: $$ x = 1 + s \cos \phi \\ y = 1 + s\sin \phi$$ and the equation of the circle is $s = 1$.