equation of the circle help?

Question

Find the equation of the circle passing through (2,3), (3,4) and (-1,2)

Answer 1

Let circle's eqn. be

x^2+y^2+2gx+2fy+c = 0

So, 4g + 6f + c +13 = 0
6g + 8f + c + 25 = 0
-2g + 4f + c + 5= 0

Solve for g,f and c.

Answer 2

(2 - a)^2 + (3 - b)^2 = r^2
(3 - a)^2 + (4 - b)^2 = r^2
(-1 - a)^2 + (2 - b)^2 = r^2
4 - 4a + a^2 + 9 - 6b + b^2 = r^2
9 - 6a + a^2 + 16 - 8b + b^2 = r^2
1 + 2a + a^2 + 4 - 4b + b^2 = r^2
4 - 4a + 9 - 6b = 9 - 6a + 16 - 8b
9 - 6a + 16 - 8b = 1 + 2a + 4 - 4b
2a + 2b = 12
20 = 8a + 4b
. a + b = 6
2a + b = 5
a = -1
b = 7
r^2 = 3^2 + 4^2 = 25
r = 5
(x + 1)^2 + (y - 7)^2 = 25
center at (-1,7), radius 5

Answer 3

The method in the previous answer is correct but I must point out that some of the positive signs should have been negative signs. Always draw the situation. This will sometimes show you that an answer you have calculated must be wrong and you go back and look for your mistake.