vasorasy8

Answered

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?

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 & Explanation

Dayana Zuniga

Expert

2022-07-04Added 16 answers

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:

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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

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

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

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

making the involvement of Pythagoras that much more clear.

enter" image="" description="" here

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

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

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

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

making the involvement of Pythagoras that much more clear.

icedagecs

Expert

2022-07-05Added 3 answers

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.

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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