# algebraic way to compute intersection of disks

Is there a pure algebraic way to calculate intersection of two disks (extended to spheres, ellipses)?

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What exactly do you want to compute? The intersection (if they do intersect non-trivially) is a region of the plane whose boundary consists of part of one circle and part of the other. – Robert Israel Sep 2 '11 at 7:17
@Robert : I think he is after analytic geometry tag rather than algebraic Geometry tag. – Arjang Sep 2 '11 at 7:58
Perhaps the Question is how to check if two disks have a nonempty intersection? This is easy for disks, not so easy for ellipses. – hardmath Sep 2 '11 at 10:36
@hardmath: No i did not mean that. Robert: Intersection of two discs would be a region that is enclosed by two arcs. Can that be categorized with just teh algebbra without any geometry. – I J Sep 2 '11 at 15:08
...so what is to be computed, then? Certainly one can shade pixels on a screen depending on whether they're inside the lens formed by the two circles or not. If you're asking about finding the intersection points of the two circles, that's easy to do. What do you need? – J. M. Sep 2 '11 at 15:29

I think what you mean is a parametric representation. Suppose the circles $C_1$ and $C_2$ intersect at points $P_1$ and $P_2$. Let $f$ be inversion around $P_1$. This maps the circles to straight lines $L_1 = f(C_1)$ and $L_2 = f(C_2)$, and the intersection of the disks to one of the four regions into which these lines divide the plane. This can be parametrized by $f(P_2) + r (\cos(\theta),\sin(\theta))$, $0 \le r < \infty$, $\theta_1 \le \theta \le \theta_2$. Invert back and you get the intersection of the disks.