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In the below figure, the point $P$ is inside the ellipse, $A, B, C, D$ are points on the ellipse, and segment $AB$ is perpendicular and intersects with segment $CD$ at $P$. The line $PE$ is perpendicular and intersects with segment $AC$ at $E$, and the line $PE$ intersects with the opposite segment $BD$ at $F$. The line $PG$ is perpendicular and intersects with segment $AD$ at $G$, and the line $PG$ intersects with the opposite segment $BC$ at $H$. The line $PI$ is perpendicular and intersects with segment $BD$ at $I$, and the line $PI$ intersects with the opposite segment $AC$ at $J$. The line $PK$ is perpendicular and intersects with segment $BC$ at $K$, and the line $PK$ intersects with the opposite segment $AD$ at $L$.

How to prove that the eight points $E, F, G, H, I, J, K, L$ are in the same circle and this circle's center and radius are only determined by point $P$'s position, in other words, the positions of $A, B, C, D$ don't influence this circle's center and radius. I have investigated this problem when the ellipse is converted to a circle, wolfram mathworld and Eight points in a circle, but I don't know whether it can help to solve the ellipse problem.

If the ellipse function is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the coordinates of point $P$ is $(s, t)$ and guarantees $\frac{s^2}{a^2} + \frac{t^2}{b^2} < 1$, give this circle's equation.

The figure of this problem

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Dandiss Valjean
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1 Answers1

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The proof can be found in Theorem 4 of Josefsson, Martin (2012), "Characterizations of Orthodiagonal Quadrilaterals" (PDF), Forum Geometricorum, 12: 13–25, where it's referred to as the second eight point circle theorem. It's too long to reproduce here, but the URL should be stable and Forum Geometricorum is published online. I don't think it addresses why the circle is independent of the positions of $A,B,C,D$, so this part of the question is still open.

The Wikipedia entry on Orthodiagonal Quadrilaterals refers to this theorem.

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