How does the Pythagorean theorem describe a circle?

Question

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?

Answer 1

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:

Owing 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

$$O = \{(a,b) \in \Bbb R^2 | \sqrt{a^2 + b^2} = c\}$$

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

$$O = \{(a,b) \in \Bbb R^2 | a^2 + b^2 = c^2 \}$$

making the involvement of Pythagoras that much more clear.

Answer 2

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.

Answer 3

Let $a$ and $b$ be the horizontal and vertical distance from the center to a point on the circle.

$a^2+b^2=r^2$ tells us that this point is $r$ away from the center.

The set of points such that they are all $r$ away from a center is a definition of the circle.

So, that means $x^2+y^2=r^2$ represents a circle.