How do I convert $(x-1)^2 + (y-\sqrt 3)^2 ≤ 4$ into polar form?

Question

I need to draw the region underneath:

$y ≤ x/\sqrt 3$

and the the circle:

$(x-1)^2 + (y-\sqrt 3)^2 ≤ 4$

My guess would be:

$x = 2cos(\theta) +1$ and $y = 2cos(\theta) +\sqrt 3$.

But the answer is $r ≤ 2cos(\theta) - 2/\sqrt 3 sin (\theta)$.

Can someone help me out?

Answer 1

Take $x =r\cos \theta, y =r\sin \theta$ and simplify.

Answer 2

Your transformation is incorrect as you are only considering points on circle with centre (1,3^1/2) and radius 2. While you need to consider the whole space hence radius must be r- one of the coordinates in polar system. The answer is given in polar coordinates with origin of given coordinates as centre. Also in answer there must be some restriction to theta to satisfy the first condition.

Answer 3

$(x-1)^2 + (y-\sqrt 3)^2 = 4$ is the equation of a circle, center $(1,\sqrt{3})$, radius$=2$.

The point $(0,0)$ is on the circle.

The straight line $y=\frac{x}{\sqrt{3}}$ cuts the circle at point A$(3,\sqrt{3})$.

On polar coordinates : $\quad\begin{cases} x=\rho\cos(\theta) \\ y=\rho\sin(\theta) \end{cases}$

Point A : $\quad\rho_A=2\sqrt{3}\quad;\quad \theta_A=\frac{\pi}{6}$.

The angle of the tangent to the circle at $(0,0)$ is $\theta_0=-\frac{\pi}{6}$.

The polar equation of the circle is :

$$(\rho\cos(\theta)-1)^2 + (\rho\sin(\theta)-\sqrt 3)^2 = 4$$ After simplification : $$\rho=4\cos(\frac{\pi}{3}-\theta)$$

The condition for a point to be on the blue area is : $$-\frac{\pi}{6}\leq\theta\leq\frac{\pi}{6}\quad\text{and}\quad 0\leq\rho\leq4\cos(\frac{\pi}{3}-\theta)$$