No you don't need to guess.
The center of the circle is a point (the only point) that is equidistant from the 3 points.
We know the equation for a circle with 3 points and we know three points on this circle. How? You guys already did the work.
You know the length of the hypotenuse. To make it easy turn the right triangle on end so that the hypotenuse is vertical and the arc is to the right. set the coordinates of the:
bottom point to 0,0
The top point to 0,(hypot)
the arc point to 1/2(hypot), 7
You have they xy coords for all 3 points.
We know the legs of the right triangle are 26.25 and using a^2+b^2=c^2 gives us the length of the hypotenuse.
So 26.25^2 + 26.26^2 = hypot ^2
or 37.123
So [0,0] [0,37.123] [7,18.56]
Now we can use the equation
(x^2+y^2-x3^2-y3^2)*(x1-x3)*(y2-y3)
+ (x1^2+y1^2-x3^2-y3^2)*(x2-x3)*(y-y3)
+ (x2^2+y2^2-x3^2-y3^2)*(x-x3)*(y1-y3)
- (x^2+y^2-x3^2-y3^2)*(x2-x3)*(y1-y3)
- (x1^2+y1^2-x3^2-y3^2)*(x-x3)*(y2-y3)
- (x2^2+y2^2-x3^2-y3^2)*(x1-x3)*(y-y3) = 0
This equation is satisfied by all points whose coordinates are (x,y)
on the circle.
Expand all that out, and gather together all terms involving x^2, x,
y^2, and y. Then complete the square on x and y to put the equation in
the form
(x-h)^2 + (y-k)^2 = r^2.
Then the center of the circle is (h,k) and the radius is r.
I am too lazy to do the math...
See
http://mathforum.org/library/drmath/view/55166.html for the formula given above. There are several other ways to go about this. You can also figure out the perpendicular bisectors, use determinants, use triangles... etc.
have fun. If it were me, I would go find the same tank and make a pattern!
It's tough to plant a brackish tank.
