vasorasy8

2022-07-03

How does the Pythagorean theorem describe a circle?
The Pythagorean theorem states, for a right triangle with legs a,b and hypotenuse c,
${a}^{2}+{b}^{2}={c}^{2}$
By replacing c with r, radius this equation becomes the equation of circle at centre (0,0).
How does Pythagoras' equation end up describing the circle?

Dayana Zuniga

Expert

It is ultimately tied to the notion of distance and how we calculate it in the xy-plane that we're familiar with. Imagine placing a right triangle, with sides a,b,c, with one vertex at the origin, like below:
enter" image="" description="" hereOwing to the dimensions, the vertices obviously lie at (a,0) and (a,b) and (0,0) (the latter by assumption of course). Then the distance from (a,b) to the origin is $\sqrt{{a}^{2}+{b}^{2}}$ by the distance formula - or, equivalently, c by construction (and the Pythagorean theorem as well).
For each point (x,y) on the circle, that distance needs to remain constant - that distance being the distance between (x,y) and (0,0). That distance is perfectly described by c - in fact, it is exactly the radius of the circle!
Imagine continuously varying a,b so that c remains constant. Then that vertex that's not on the horizontal axis ultimately traces out a circle as a result. We could define this circle O by
$O=\left\{\left(a,b\right)\in {\mathbb{R}}^{2}|\sqrt{{a}^{2}+{b}^{2}}=c\right\}$
to establish the whole "distance to the origin remains constant" thing: after all, that's the defining property of a circle, the set of points equidistant from a given point (here, the origin). Equivalently, though, we see by squaring both sides of that latter equality
$O=\left\{\left(a,b\right)\in {\mathbb{R}}^{2}|{a}^{2}+{b}^{2}={c}^{2}\right\}$
making the involvement of Pythagoras that much more clear.

icedagecs

Expert

This circle has the origin as it centre, so the four Cartesian quadrants each contain a quarter of it. In the first quadrant, a,b are both positive. Consider the right-angled triangle whose vertices are (0,0),(a,0),(a,b) with ${a}^{2}+{b}^{2}={c}^{2}$ by Pythagoras; the third vertex lies on the circle, so smoothly rotating the length-c hypotenuse from (c,0) to (0,c) traces out the quarter-circumference in the first quadrant.
As we continue through other quadrants, the right-angled triangle ends up flipped horizontally and/or vertically, but the above formulae for its vertices remain correct. In short, we make a circle by rotating a radius, and each location it has along the way lets us construct such a triangle.

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