Write the general equation for the circle that passes through the points: (1, 7) (8, 6) (7, -1)
The three points are: A(1,7), B(7,-1) and C(8,6).
The slope of BC is m1 = 7.
Mid-point of BC is D = (15/2, 5/2).
Slope of perpendicular line through D is m2 = -1/m1 = -1/7.
Perpendicular line through D is:
(y – 5/2)/(x – 15/2) = -1/7
x + 7y = 25. ------------ (1)
This line is a perpendicular bisector of a chord of a circle and, as such, passes through the centre of the circle.
The slope of AB is m3 = -4/3.
Mid-point of AB is E = (4, 3).
Slope of perpendicular line through E is m4 = -1/m3 = 3/4.
Perpendicular line through D is:
(y – 3)/(x – 4) = 3/4
3x - 4y = 0. -------------- (2)
This line is a perpendicular bisector of a chord of a circle and, as such, passes through the centre of the circle.
The intersection of lines (1) and (2) give the centre of the circle.
Using x = 25 – 7y, from (1),
3(25 – 7y) – 4y = 0
75 = 3y
y = 3, x = 4
The centre point of the circle is P(4,3)
The radius is given by distance from centre to a point on the circle, i.e. R = AP.
Where A = (1,7) and P = (4,3)
R = sqrt([1 – 4]^2 + [7 – 3]^2) = sqrt(3^2 + 4^2) = sqrt(5^2) = 5
R = 5
The equation of the circle is then:
(x – 4)^2 + (y – 3)^2 = 25
x^2 + y^2 - 8x - 6y = 0