For example, another point on our circle is (3/√2, 3/√2). You will find that this works for every single point on the circle. Using this in the Pythagorean theorem, we find:ĭoes this work for the point we selected (aka (3, 0))? The hypotenuse would be the radius of our circle. The base of the triangle would be the x-axis, and the adjacent side would be some y-value.
Since this is a right triangle, we should be able to apply the Pythagorean theorem. Let's try to make a right triangle, where the center of the circle is one vertex, and its opposite vertex is the outer edge.
Let's find some points on the outer edge of the circle.Ī noticeable one is (3, 0) (3 units away from the center). Let's consider a circle with center (0, 0) (to make the explanation a little simpler) and radius 3.
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