# 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?

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.

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.

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.