### No Calculus. No Algebra. Just Plane Geometry.

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

**. The circle touches the parabola at exactly two points. What are the coordinates of the centre of the circle?**

*1***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.

Mark as ** F** the lower point of intersection between the circle and the vertical line. Construct a circle through

**with**

*C***as the centre. The two circles share the radius**

*F***. Therefore they both have a radius of**

*CF***.**

*1*Mark the points where the two circles intersect: ** P** and

**. Construct a third circle through**

*P’***and**

*C***with a centre at**

*F***. This circle also has a unit radius.**

*P*Find the intersection (other than ** C**) between circles

**and**

*P***. Label this point**

*F***and connect points**

*D***and**

*P***. Segment**

*D***will have a unit length.**

*PD*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__.)

Mark 3 more points: The second intersection between ** F** and the vertical line,

**; The intersection between the two straight lines,**

*C’***and the midpoint between**

*F’;***and**

*F*

*F’*,**(**

*V**)*

__Euclid 1.10__**.**

Since ** FDC’** are the vertices of an equilateral triangle (

*),*

__Euclid 1.1__**is the midpoint of**

*F’***Therefore Segment**

*FC’.***has a length of**

*CV***.**

*5/4*Finally, consider a parabola with ** V** as the vertex and

**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***. Since**

*F’D***and**

*PF***are both of unit length, point**

*PD***must be on the parabola. By a similar argument, so is point**

*P*

*P’.*It remains to place the problem in its original context, a parabola and circle on the Cartesian plane.

With the vertex at the origin, the centre of the circle, ** C** is at

**(0, 5/4)**. Points

**and**

*P***are vertices of equilateral triangles with unit sides. Thus, they each have a distance of**

*P’***√3/2**from the

**axis**

*y-***.**Each has a

*y-*component

*1/2***l**ess than that of

**That sets their coordinates at**

*C*.**and**

*(√3/2, 3/4)***, respectively**

*(-√3/2, 3/4)***.**

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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