Equation of a circle in polar coordinates under a linear transformation

Question

Let's say we translate a circle with origin $(0,0)$ on the x axis by some constant $c$. What would the new equation of the circle be in polar coordinates? I have tried subbing in the equation of the translated circle $f(x-c)$ but I am not sure how to convert it into a function in polar coordinates (in terms of $r$ and $\theta$).

Answer 1

The Cartesian equation would be $(x-c)^2+y^2=x^2+y^2+c^2-2xc=R^2$, where $R$ is the radius of the circle. Letting $x = r\cos(\theta), y = r\sin(\theta)$, we have $$R^2 = r^2+c^2-2rc\cos(\theta)\implies$$ $$r^2-r\cdot(2c\cos(\theta))+(c^2-R^2)=0 \implies$$ Solving for $r$ by using the quadratic formula and reducing, $$r = c\cos(\theta)+\sqrt{R^2-c^2\sin^2(\theta)}$$ You might notice that the formula doesn't work if $c > R$. This is because the origin would not be inside of the circle, and so some $\theta$ values would not correspond to any $r$ value, and some would correspond to two.