F.5 maths locus

This question has alternative solution.
Let A= (-1,2) , B=(3,2) , C= (4,-3) , Centre of the circle be M = (a,b)
Actually AB, BC, AC are the chords of the circle.
By using (line joining from centre to mid pt of chord perpendicular chord)
Let mid pt of AB be X = [(-1+3)/2 , (2+2)/2) = (1,2)
MX perpendicular to AB,
AB is a horizontal line y=2,
it implies MX is a vertical line.
The x-coordinates of M = The x-coordinates of X = 1
So M=(1,b)
Let mid pt of BC be Y = [(3+4)/2, (2-3)/2] = (7/2 , -1/2)
MY perpendicular to BC,
Slope of MY x Slope of BC = -1
[(b+1/2)/(1-7/2)] x [(2+3)/(3-4)] = -1
Hence b = -1
Centre of the circle = (1,-1)
Radius of the circle = sq rt [(1+1)^2+(-1-2)^2] = sq rt (13)
Equation of the circle : (x-1)^2 + (y+1)^2 = [sq rt (13)]^2
General form ->x^2+y^2-2x+2y-11= 0

But not all kinds of question can use this method!!

Actually you can ALWAYS solve this type of questions by finding the intersection point of any two perpendicular bisector, that gives the center.

The distance between the center and any point is the radius - now you have the circle!

Let the center of the circle be (a, b), so
(a + 1)² + (b - 2)² = (a - 3)² + (b - 2)²
==> a² + 2a + 1 + b² - 4b + 4＝a² - 6a + 9 + b² - 4b + 4==> 8a = 8==> a = 1 ⋯⋯⋯⋯⋯⋯⋯ (1)
(a + 1)² + (b - 2)² = (a - 4)² + (b + 3)²
==> a² + 2a + 1 + b² - 4b + 4＝a² - 8a + 16 + b² + 6b + 9==> 10a - 10b - 20 = 0==> a - b - 2 = 0 ⋯⋯⋯⋯ (2)

Sub. (1) into (2), get b = -1∴ center of the circle is (1, -1)

The equation of the circle is :(x - 1)² + (y + 1)² = (3 - 1)² + (2 + 1)²==> x² - 2x + 1 + y² + 2y + 1 = 4 + 9==> x² + y² - 2x + 2y - 11 = 0