Question

What is the center of the circle that you can circumscribe about a triangle with vertices A(2, 6), B(2, 0), and C(10, 0)?

Answer 1

The three points A,B,C are all points on this circle.
Each point is then equal distance from the center, that distance being the radius of the circle.
Using the distance formula, we can find the center of the circle  (x,y):
$d^2 = (x-x_0)^2 + (y-y_0)^2$
Plugging in points A and B into distance formula, then setting them equal to each other gives:
$(x-2)^2+(y-6)^2 = (x-2)^2 + y^2$
Right away we can cancel out the x terms leaving:
$(y-6)^2 = y^2$
Expand Left side and Solve for y:
$y^2 -12y +36 = y^2$
$y = 3$
Plug in points B and C as before:
$(x-2)^2 + y^2 = (x-10)^2 + y^2$
Here we can cancel the y-terms.
Expand and solve for x:
$x^2 -4x+4 = x^2 -20x+100$
$16x = 96$
$x = 6$
Therefore the center of the circle is the point (6,3)