Find the equation nd a circumcircle whose testices are A (2, 3), B (5, 4), and C (3, 7)
Testices ??? lol
AB and BC would be chords in your circle
and the centre would be the intersection of the right-bisectors.
One of them:
slope AB = (1/3)
so slope of perpendicular = -3
midpoint of AB = (7/2 , 7/2)
equation:
y - 7/2 = -3(x - 7/2)
y - 7/2 = -3x + 21/2
3x + y = 14
the other:
slope BC = 3/-2
slope of perp. = 2/3
midpt of BC = (4, 11/2)
equation:
y - 11/2 = (2/3)(x-4)
y - 11/2 = (2/3)x - 8/3
times 6
6y - 33 = 4x - 16
4x - 6y = -17
first one times 6
18x + 6y = 84
add them:
22x = 67
x = 67/22
I will let you confirm that y = 107/22
circle:
(x - 67/22)^2 + (y - 107/22)^2 = r^2
plug in (2,3)
(2 - 67/22)^2 + (3 - 107/22)^2 = r^2
r^2 = (-23/22)^2 + (-41/22)^2
= 2210/484
(x - 67/22)^2 + (y - 107/22)^2 = 2210/484
They could have planned this question out a bit better.
Confirm this by subbing in the original three points.