This is just a question I have been thinking about for a few days. Let's say we have a right triangle with vertices at (0,0), (a,0) and (a,b). Now, consider that an arc has been constructed with centre (x,y) and that the arc passes through the vertices (0,0) and (a,b) so the arc is 'over' the hypotenuse. What is the locus of points of the centre of the arc? Geometrically, does the centre of the arc lie on a circle/ellipse or something else?

I hope this question has made some sense - please comment if you require any further explanation!

Thank you!

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    $\begingroup$ So far, my best idea is that the centre of the circle lies on a straight line which is the line that bisects the points (0,0) and (a,b) $\endgroup$ – Dinglevak Jun 15 at 23:16

You are correct with your idea stated in the comment, i.e., the line bisects the chord of $(0,0)$ to $(a,b)$, except you didn't mention the line is perpendicular to this chord. This is because the equal length (i.e., the radius) lines from the center of the circle to each end point of this chord, plus the chord line itself, form an isosceles triangle. Thus, the perpendicular bisector of the chord goes through the opposite triangle end-point, i.e., the center of the circle.


Adding to John Omielan's answer, according to the perpendicular bisector theorem, if a point is on the perpendicular bisector, then it is equidistant from the segment's endpoints.

The converse of this theorem shows that if the point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector. Since $(x,y)$ is the centre of the circle passing through $(0,0)$ and $(a,b)$, then the centre also must lie on the perpendicular bisector.


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