Equation of off centre circle in polar coordinates

Question

Please help me understand what I'm missing here

The goal is to find an expression for r as a function of $\theta$ in the figure.

Start with the sines rule in triangle ABC $$\frac{\rho}{\sin(\theta)} = \frac{d}{\sin(\pi - (\theta +\alpha))} = \frac{r}{\sin(\alpha)}$$

using the first equality we can get $$\alpha = \arcsin(\frac{d}{\rho}\sin(\theta)) - \theta$$

substituting $\alpha$ in the last term results in $$r = \rho \frac{\sin(\arcsin({d \over \rho}\sin(\theta)) - \theta)}{\sin(\theta)}$$ expanding the numerator and cancelling $\sin(\theta)$ gives: $$r = d\cos(\theta) - \rho\sqrt{1 - \frac{d^2}{\rho^2}\sin(\theta)^2}$$ This is where it goes wrong. keeping $(d - \rho)$ constant, the limit of this expression at very large d becomes $$ \lim_{d \to \infty} r = (d-\rho)\cos(\theta)$$ where from the figure r is clearly always bigger than $d - \rho$ and at the limit should be $$\frac{d - \rho}{\cos(\theta)}$$ I've checked everything multiple times (doesn't mean it's not wrong) manually and with Matlab's symbolic math toolbox.

Thanks

Answer 1

First let's use the cosine law to get an alternative of your result. $$\rho^2=r^2+d^2-2dr\cos\theta$$ We solve for $r$, and we choose the lower value (otherwise you get the other intersection with the circle): $$r^2-2dr\cos\theta+d^2-\rho^2=0$$ which yields $$r=d\cos\theta-\sqrt{d^2\cos^2\theta-(d^2-\rho^2)}$$ This is the same formula you got, but I will not transform the cosine to sine.

You have that $d-\rho=c$. Then $$d^2-\rho^2=(d-\rho)(d+\rho)=c(2d-c)$$ Therefore $$r=d\cos\theta\left(1-\sqrt{1-\frac{c(2d-c)}{d^2\cos^2\theta}}\right)$$ Now let's use an approximate for the square root::$$\sqrt{1+x}\approx1+\frac x2$$ Therefore $$r=d\cos\theta\left(1-1+\frac{c(2d-c)}{2d^2\cos^2\theta}\right)=\frac{2dc-c^2}{2d\cos\theta}$$

Ignoring the $c^2$ term, and simplifying by $2d$, you get $$r=\frac c{\cos\theta}=\frac{d-\rho}{\cos\theta}$$

The reason it did not work for you is that you neglected the fact that the error in the square root gets multiplied by a large number ($\rho$).

Answer 2

The quadratic equation $\rho^2=r^2+d^2-2dr\cos\theta$ solutions supply three solutions/intersection points in polar coordinates depending on whether point A is outside the eccentric circle, inside or on it.

Point A outside the circle, between two tangents:

$$ r= d \cos \theta \pm \sqrt{\rho^2-(d\sin \theta)^2}$$

Point A inside the circle:

$$ \text{with}~ \alpha= \sin^{-1} \dfrac{\rho}{d}~,~ {r}= d \cos \theta \pm \rho\sqrt {1-\left(\frac{\sin \theta}{\sin \alpha}\right)^{2}}$$

A on the circle $$ r= d \cos \theta. $$

Three sketches can be drawn making the three cases clearly demarcated. The $\pm$ symbol then denotes plus and minus.