Cartesian to Polar Coordinate Transformation of a shifted circle

Question

Given the Region:

$$\displaystyle D=\ \left\{( x,y) \ \in \mathbb{R}^{2} \ |\ x^{2} \ +\ y^{2} \ \leqslant ax,\ a\ \in \mathbb{R^+}\right\}$$

We have the equivalent:

$$ \displaystyle D' =\ \left\{( r,\theta ) \in \mathbb{R}^{2} \ \middle|\ 0\leqslant r\leqslant a\cos\theta ,\ -\frac{\pi }{2} \leqslant \theta \leqslant \frac{\pi}{2}\right\}$$

The region $D$ is a circle with its center located at $\big(\frac{a}{2},0\big)$. So from the graph of $D$ someone can understand that $-\frac{\pi}{2} \leqslant \theta \leqslant \frac{\pi}{2} $.

Can someone explain the transformation?

Answer 1

$x^2+y^2\le ax$ is equivalent to $r^2\le ar\cos{\theta}$

Subtracting $ar\cos{\theta}$ from both sides of the inequality gives $r^2-ar\cos{\theta}=r(r-a\cos{\theta})\le 0$.

Since we require $r$ to be non-negative it follows that the inequality is true only when $r-a\cos{\theta}\le 0$.

That's how we get the inequality $0\le r\le a\cos{\theta}$.

To understand why the angle is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ instead of $-\pi$ and $\pi$ you have to sketch the circle using polar coordinates. If you go from $0$ to $2\pi$ you will traverse the circle twice.

Answer 2

The circle is given by

$$ \left(x-\frac{a}{2}\right)^2 + y^2 = \frac{a^2}{4} $$

It must be true that

$$ -\frac{a}{2} \le x - \frac{a}{2} \le \frac{a}{2} $$

or

$$ 0 \le x \le a $$

Hence, the circle is entirely on the right half-plane (if $a > 0$ that is)