# Pythagoras theorem is a^2 + b^2 = c^2 and a circle has an equation x^2 + y^2 = a^2 .Is there a relation between a right angle triangle and a circle?

I was just curious about the fact that whether such a relation exists when I came across the equation of a circle.(I maybe absolutely wrong) .

• Afaik. Sometimes trigonometric functions are defined in terms of a unit circle. You can check it out. May 28, 2019 at 18:00
• See this video to understand why the answer to your question is basically: YES. May 28, 2019 at 18:04
• Here is another animation that shows this. May 28, 2019 at 18:06
• xy coordinate system.A(-c/2,0); B(0,c/2),C(x,y).Consider locus C(x,y): $x^2+y^2=(c/2)^2$.We have BC=a;AC=b.Thales circle $a^2+b^2=c^2$. May 28, 2019 at 18:23
• Any circle diameter and any circle point connected to the diameter's ends forms a right angle triangle having the diameter as a hypotenuse. So definitely there is a relation between circle and right angle triangle.
– Nick
May 28, 2019 at 18:47

Yes they're related! A circle is the locus of a point, which is always equidistant from the center.

Now $$P$$ is a point on the circle of radius $$r$$ with co-ordinates $$(x,y)$$.

By Pythagoras theorem

$$r^2=x^2+y^2$$

Which is the equation of the circle!

If we have a segment $$AB$$ and $$O$$ is the midpoint, a circle is formed by all possible locations for the third vertex of a right triangle that has $$AB$$ as the hypotenuse and $$O$$ will be the center of the circle.

This relationship can be summed up as:

For any right triangle there exists only one semi-circle whose diameter is equal hypotenuse of the right triangle.

Conversely: For any semi circle there exists infinitely many right triangles whose hypotenuse is equal to the diameter of the semi-circle