# Number of Intersection points between lower half of an ellipse and a circle

An ellipse has its axes parallel to the coordinate system axes and its major axes is parallel to X-axis.

Meanwhile, there is a circle located at the coordinate system origin, whose radius is smaller than the semi major of the ellipse.

Now a curve is constructed as the lower half of the said ellipse.

How do I find out how many intersection points will exist between the circle and the curve? The general case can be 1 point, 2 points and 3 points (shown in the figures below). Is it possible to have 4 intersection points?

• How do you have 3 intersections when the curve is only the lower half of the ellipse? – Lee David Chung Lin Nov 1 '18 at 15:50
• It happens when the half ellipse has the lowest point tangent to the circle and the curve curves up. – Steph Fong Nov 1 '18 at 17:18
• Ah, I would like to ask about this curvature. The curvature of the ellipse is constantly changing. I guess we have to prove that the curvature is always smaller than the circle? – Steph Fong Nov 1 '18 at 18:03
• No, I don't see 3 pionts of intersection on the plot, unless the minor axis is along the X direction. I would like to send you a picture, but I don't know how to do it with "comment". If you don't mind, I do like to get technical on this. So to prove that in a neighborhood around the lowest point there will be no other intersection points, we have to first calculate the curvature of the curve and examine if in [-r, r], where r is the radius of the circle, the curvature is always smaller than the curvature of the circle, right? – Steph Fong Nov 1 '18 at 18:30
• @Lee Three intersections on the lower half is perfectly possible. I just did it in Geogebra. The measurements from Geogebra (unfortunately only with $2$ decimals) are: $r=4$, foci of the ellipse $(7,-3)$ and $(-2.59,-2.99)$, and the semi-major is $4.85$. – Jens Nov 1 '18 at 18:34

Yes, you can have 4 intersection points. Take for instance four points on the half-ellipse symmetric around the minor axis and the circle passing through them. But non-symmetric solutions are also possible.

• The semi major of the ellipse should be larger than the radius of the circle. – Steph Fong Nov 1 '18 at 23:26
• In the first version of your question you wrote that the major axis of the ellipse is larger than the radius of the circle. I'll think about your new request. – Aretino Nov 2 '18 at 7:20

Your situation is of intersection between two conics of second order. Let us consider the total number r of points of intersection comprehensively.

Apart from zero number of intersections for non-intersecting complex case, you can only have an even number of real intersection points either two or four, while solving for their solutions.

In tangential case these become correspondingly one or two double points with discriminant vanishing for a double root.

You cannot have odd number of real intersections no matter where you draw them in the $$(x,y)$$ plane. If you draw your right figure fully you would encounter one more intersection at right in your diagram.

• I would like to study the lower half of the ellipse only and also the situation where the semi major radius of the ellipse is larger than the radius of the circle. – Steph Fong Nov 1 '18 at 23:28
• I dont get what you want. There can be only four four real intersections. You sketched correctly three of them at left, excluding the fourth point, then zooming it at right in a separate sketch including the fourth What else now do you want to study or know next? A table of $(x,y)$ co-ordinates numerically? – Narasimham Nov 2 '18 at 6:08
• @Narasimham I made the sketch, and if that misled you I apologize. The zoomed-in picture is also showing only 3 intersections. Where do you see the fourth? The other marked point is the center of the ellipse, with the dotted lines indicating the axes are aligned. – Lee David Chung Lin Nov 2 '18 at 12:58
• @Steph Fong By the way, feel free to edit over my edit if you don't like it (the picture, the typesetting, etc). You can always do that because you own the question post. Having said that, one must wonder why so many people (including myself) misunderstand your question on the first read. Maybe more things need be changed in the question text to put some emphasis on the key points. – Lee David Chung Lin Nov 2 '18 at 13:01
• Hope you see the fourth point that is lost/excluded in the smaller one as you chose a frame outside of the fourth point 4, I labelled your diagrams with the three points in the zoom and all four points included in the smaller one at left. Hope its OK. – Narasimham Nov 2 '18 at 16:46