Find the circumcircle of a triangle with vertices (0,0)(0, 0)(0,0), (4,0)(4, 0)(4,0), and (0,3)(0, 3)(0,3).
(x−2)2+(y−32)2=254(x - 2)^2 + (y - \frac{3}{2})^2 = \frac{25}{4}(x−2)2+(y−23)2=425
(x−2)2+(y−1.5)2=5(x - 2)^2 + (y - 1.5)^2 = 5(x−2)2+(y−1.5)2=5
x2+y2−4x−3y=0x^2 + y^2 - 4x - 3y = 0x2+y2−4x−3y=0
x2+y2=25x^2 + y^2 = 25x2+y2=25