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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) .

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  • $\begingroup$ Afaik. Sometimes trigonometric functions are defined in terms of a unit circle. You can check it out. $\endgroup$
    – Someone
    May 28, 2019 at 18:00
  • $\begingroup$ See this video to understand why the answer to your question is basically: YES. $\endgroup$
    – Crostul
    May 28, 2019 at 18:04
  • $\begingroup$ Here is another animation that shows this. $\endgroup$
    – Bram28
    May 28, 2019 at 18:06
  • $\begingroup$ 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 $a^2+b^2=c^2$. $\endgroup$ May 28, 2019 at 18:23
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    $\begingroup$ 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. $\endgroup$
    – Nick
    May 28, 2019 at 18:47

3 Answers 3

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

enter image description here

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!

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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.

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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

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