TITLE: Ellipse and circle
QUESTION [3 upvotes]: if $\alpha$, $\beta$, $\gamma$, $\delta$ be the eccentric angles of four points of intersection of the ellipse and any circle,prove  that $\alpha+\delta+\beta+\gamma$ is an even multiple of $\pi$ radians.

REPLY [3 votes]: WLOG the equations of the Circle & the Ellipse  can be chosen to be $$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\text{ and } x^2+y^2=r^2$$
We know the parametric coordinate of any $P(x,y)$ on the ellipse is $(h+a\cos\phi,k+b\sin\phi)$ where $\phi$ is the eccentric angle
To find the intersection, $$(h+a\cos\phi)^2+(k+b\sin\phi)^2=r^2$$
Use  Weierstrass substitution, to form a Bi-Quadratic Equation in $\tan\frac\phi2$ whose roots are $\displaystyle\tan\frac\alpha2,\tan\frac\beta2,\tan\frac\gamma2,\tan\frac\delta2$
Now using $\tan(A+B+C+D)$ formula , 
$$t^4\{(h-a)^2+k^2-r^2\}+(4bk)t^3+()t^2+(4bk)t+()=0$$
Using Vieta's formula  $$\sum \tan\frac\alpha2=-\frac{4bk}{(h-a)^2+k^2-r^2}=\sum \tan\frac\alpha2\tan\frac\beta2\tan\frac\gamma2$$
$$\implies\tan\left(\frac{\alpha+\beta+\gamma+\delta}2\right)=0\implies \frac{\alpha+\beta+\gamma+\delta}2=n\pi$$ where $n$ is any integer
{"set_name": "stack_exchange", "score": 3, "question_id": 623231}