I want to prove that this polar equation:
$$r^2 + 2r(\cos(\theta) – 3\sin(\theta)) = 4$$
describes a circle.
I tried converting the equation into a cartesian equation and got
$$r^2 + 2x – 6y = 4$$
and this is basically where i got stuck. Where do I go from here, or am I proving this in the correct way at all?
Best Answer
A circle is described in polar form by $r^2-2rx_0\cos(\theta - y_0)+x_0^2=a^2$ where $a$ is a constant and $x_0$ and $y_0$ represent the coordinates of the center of the circle. Therefore, $$\frac{dr}{d\theta}=2rx_0\sin(\theta-y_0)$$ So if the equation you are given describes a circle, then $\frac{dr}{d\theta}=2rx_0\sin(\theta-y_0)$. All you have to do is prove this. Implicit differentiation will be needed, though.