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copied!<p>So the solution is here:</p> <p>The parametrized formula of an ellipse:</p> <pre> x = x0 + a * cos(t) y = y0 + b * sin(t) </pre> <p>Let's put known coordinates of two points to it:</p> <pre> x1 = x0 + a * cos(t1) x2 = x0 + a * cos(t2) y1 = y0 + b * sin(t1) y2 = y0 + b * sin(t2) </pre> <p>Now we have a system of equations with 4 variables: center of ellipse (x0/y0) and two angles t1, t2</p> <p>Let's subtract equations in order to get rid of center coordinates:</p> <pre> x1 - x2 = a * (cos(t1) - cos(t2)) y1 - y2 = b * (sin(t1) - sin(t2)) </pre> <p>This can be rewritten (with product-to-sum identities formulas) as:</p> <pre> (x1 - x2) / (2 * a) = sin((t1 + t2) / 2) * sin((t1 - t2) / 2) (y2 - y1) / (2 * b) = cos((t1 + t2) / 2) * sin((t1 - t2) / 2) </pre> <p>Let's replace some of the equations:</p> <pre> r1: (x1 - x2) / (2 * a) r2: (y2 - y1) / (2 * b) a1: (t1 + t2) / 2 a2: (t1 - t2) / 2 </pre> <p>Then we get simple equations system:</p> <pre> r1 = sin(a1) * sin(a2) r2 = cos(a1) * sin(a2) </pre> <p>Dividing first equation by second produces:</p> <pre> a1 = arctan(r1/r2) </pre> <p>Adding this result to the first equation gives:</p> <pre> a2 = arcsin(r2 / cos(arctan(r1/r2))) </pre> <p>Or, simple (using compositions of trig and inverse trig functions):</p> <pre> a2 = arcsin(r2 / (1 / sqrt(1 + (r1/r2)^2))) </pre> <p>or even more simple:</p> <pre> a2 = arcsin(sqrt(r1^2 + r2^2)) </pre> <p>Now the initial four-equations system can be resolved with easy and all angles as well as eclipse center coordinates can be found.</p>

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