SOLUTION: So there is a circle with a radius of ten and the center of the circle at (10,10) and the circle is tangent to the x and y axis. There is an equilateral triangle inscribed to the c

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Question 877498: So there is a circle with a radius of ten and the center of the circle at (10,10) and the circle is tangent to the x and y axis. There is an equilateral triangle inscribed to the circle with one of the vertex's at (20,10). How do you find the length of one side of the triangle? How do you find the coordinates of the other vertices?
Thank you for your time and help!!
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
So there is a circle with a radius of ten and the center of the circle at (10,10) and the circle is tangent to the x and y axis. There is an equilateral triangle inscribed to the circle with one of the vertex's at (20,10). How do you find the length of one side of the triangle? How do you find the coordinates of the other vertices?
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The radius of the triangle = 10
Lines from the center to the 3 vertices make 3 isoceles triangles.
Lines from the center perpendicular to the 3 sides make 6 right triangles with hyp = 10 and angles of 30 degrees.
The perpendicular lines from the center are 5 units long.
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Pythagoras --> 1/2 the side of the triangle = 5sqrt(3) units.
Each side = 10sqrt%283%29
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The side parallel to the x-axis is 5 units from the center --> y = 5
The circle is %28x-10%29%5E2+%2B+%28y-10%29%5E2+=+100
Sub 5 for y
%28x-10%29%5E2+%2B+%285-10%29%5E2+=+100
x%5E2+-+20x+%2B+25+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-20x%2B25+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-20%29%5E2-4%2A1%2A25=300.

Discriminant d=300 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--20%2B-sqrt%28+300+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-20%29%2Bsqrt%28+300+%29%29%2F2%5C1+=+18.6602540378444
x%5B2%5D+=+%28-%28-20%29-sqrt%28+300+%29%29%2F2%5C1+=+1.33974596215561

Quadratic expression 1x%5E2%2B-20x%2B25 can be factored:
1x%5E2%2B-20x%2B25+=+%28x-18.6602540378444%29%2A%28x-1.33974596215561%29
Again, the answer is: 18.6602540378444, 1.33974596215561. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-20%2Ax%2B25+%29

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x+=+10+%2B-+5sqrt%283%29
Vertices at (1.34,5) and (18.66,5)