The Circle-Parabola Problem

Adam Hrankowski

No Calculus. No Algebra. Just Plane Geometry.
Actual image of Euclid solving a problem he found on InstagramWikipedia Commons

In June 2021, Dr. Peyam (of YouTube fame) published a video of a geometry problem he found on Instagram. The solution, he says, “uses a lot of beautiful geometry and even more beautiful calculus.” (It does indeed. Check it out.)

I am going to attack the same problem using methods that predate calculus — and even algebra. Here goes!

The Problem

A parabola formed by the graph of y=x² cradles a circle of radius 1. The circle touches the parabola at exactly two points. What are the coordinates of the centre of the circle?
Illustration by Author

My Approach

We’ll solve this by constructing the diagram, using only a compass and straight-edge. (And Geogebra.)

Begin with a vertical line. Find a point on it, C. Draw a circle around that point. We’ll define the radius of the circle as our unit.
Illustration by Author

Mark as F the lower point of intersection between the circle and the vertical line. Construct a circle through C with F as the centre. The two circles share the radius CF. Therefore they both have a radius of 1.
Illustration by Author

Mark the points where the two circles intersect: P and P’. Construct a third circle through C and F with a centre at P. This circle also has a unit radius.
Illustration by Author

Find the intersection (other than C) between circles P and F. Label this point D and connect points P and D. Segment PD will have a unit length.

Construct a line through D tangent to circle P and perpendicular to the vertical line. (This is not a primitive construction, like drawing a circle or a straight line. You can find the detailed construction in Euclid 1.11.)
Illustration by Author

Mark 3 more points: The second intersection between F and the vertical line, C’; The intersection between the two straight lines, F’; and the midpoint between F and F’, V (Euclid 1.10).
Illustration by Author

Since FDC’ are the vertices of an equilateral triangle (Euclid 1.1), F’ is the midpoint of FC’. Therefore Segment CV has a length of 5/4.

Finally, consider a parabola with V as the vertex and F as its focus. A parabola is that set of points that are equidistant from a focus, F, and a given straight line, the directrix, F’D. Since PF and PD are both of unit length, point P must be on the parabola. By a similar argument, so is point P’.
Illustration by Author

It remains to place the problem in its original context, a parabola and circle on the Cartesian plane.
Illustration by Author

With the vertex at the origin, the centre of the circle, C is at (0, 5/4). Points P and P’ are vertices of equilateral triangles with unit sides. Thus, they each have a distance of √3/2 from the y-axis. Each has a y-component 1/2 less than that of C. That sets their coordinates at (√3/2, 3/4) and (-√3/2, 3/4), respectively.

These points establish the parabola as y=x². The circle is x²+(y-5/4)²=1

Which approach do you prefer?

[Originally published June 7, 2021 on Medium]

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I thought I was a renaissance man. Turns out I have ADHD.


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