Thread: A circle contains two vertices of a square and is tangent to the opposite side.

A circle contains two vertices of a square and is tangent to the opposite side. If each side of the square has length 10, find the radius of the circle. Write your answer as a simplified common fraction.

Could I get an answer, and if possible, a picture?

Could you state the whole question word for word?

A circle contains two vertices of a square and is tangent to the opposite side.

What is tangent to what now? This is the part you need to clarify.

nvm, that was the whole question, but I got it. the two verticies are points on the circle, and the side opposite the vertices is tangent to the circle.

Originally Posted by help1

nvm, that was the whole question, but I got it. the two verticies are points on the circle, and the side opposite the vertices is tangent to the circle.

Ok, so the whole square is not inside the circle?

Makes sense now.

Do you still need me to help?

Consider the square with vertices: (0,0), (10,0), (10,10), & (0,10).
Now you should know how to find the equation of a circle determined by three non-collinear points. So do that for the points (0,0), (0,10), & (10,5).
Actually it made sense the first time.

Let the radius of the circle be x. If the square is tangent to the circle at T and the vertices A and B are on the circumfernce.



Solve.

(10-x)^2 + 5^2 = x^2

Thanks guys!