# Can we construct an intersection of an ellipse and a circle sharing the same center when the center and five points on an ellipse is given? [closed]

I am given: Five points on the ellipse, The sharing center of the ellipse and the circle, And the circle.
One of those five points is one of the intersections of a circle and an ellipse.
Can we construct the three intersections of an ellipse and a circle that is not given?

Blue ones are given, and greens are the points I want to find.

• Sure. All you need to do is solve the equation of the circle (or the ellipse) for x or y, substitute the result into the other equation and solve for the remaining variable. Once you have values for it, you can use either equation to solve for the remaining one. You will end up with a list of the coordinates of the intersections. Commented Jun 17, 2023 at 0:36
• I mean the Euclidean construction, where we only use the ruler and compass to draw. Is there any geometrical way to do so? Commented Jun 17, 2023 at 1:02
• the most detail on constructions is in Heinrich Dorrie, 100 Great Problems of Elementary Mathematics. In section 64, pages 273-274 especially, he says how to take a conic section, defined by five points, and find its intersection with another line. This is basically Pascal's Hexagon Theorem, section 61, pages 257-261 Commented Jun 17, 2023 at 1:34
• @WillJagy I've found out that intersection of a conic and a straight line is able to be constructed, but I am looking for a construction of the intersection of a conic and a circle. I looked out the book you mentioned but failed to find the construction of the intersection of a conic and a circle. Commented Jun 17, 2023 at 4:17
• As I explained in my answer, all you have to do is to construct the axes of the ellipse. Another construction for the axes can be found here: math.stackexchange.com/a/4393709/255730 Commented Jun 17, 2023 at 10:05

Let $$ABCDE$$ be the five given points on the ellipse. We can use Pascal's theorem to find the line tangent to the ellipse at $$A$$: if $$F$$ is the intersection of $$AB$$ with $$CD$$, and $$G$$ is the intersection of $$AC$$ with $$BE$$, then the intersection $$T_A$$ of $$FG$$ with $$DE$$ is a point on the tangent at $$A$$ to the conic $$ABCDE$$ (see here for a detailed proof).
Draw through center $$O$$ the parallel $$p$$ to the tangent at $$A$$ and consider another point $$B$$ on the ellipse. Let $$H$$ be the intersection between line $$OA$$ and the parallel to $$p$$ through $$B$$, $$K$$ be the intersection between line $$p$$ and the parallel to $$OA$$ through $$P$$.
You can then construct point $$A'$$ on line $$p$$, such that $$OA'$$ is a conjugate semi-diameter of $$OA$$, finding $$OA'$$ from the equality (Apollonius equation for the ellipse): $$\left({OK\over OA'}\right)^2+\left({OH\over OA}\right)^2=1.$$