What is the equation of the circle?

Consider the three given points
A( -2, 4), B( 6, 0), C( 6, 8)

as the verteces of a triangle, and
a = BC, b = AC, c = AB

as its sides, which are chords of the circumcircle of this triangle.

The centre point M of the circumcircle is defined as the intersection point of the triangle's perpendicular side bisectors.

One such bisector is already given with the condition: "if the diameter is perpendicular to a chord (which is not a diameter) whose endpoints are B(6,8) and C(6,0)".

A diameter by definition passes through the centre point M, and the perpendicular bisector of each possible chord passes through M.
As a=BC is vertical ( xB = xC = 6 ), the perpendicular bisector of side a, a', has to be horizontal. As it goes through A(-2,4) it has to bisect side a in Pa( 6, 4). Therefore,
yM = 4, and

a' := y = 4.

Now we have to find the bisecting point Pc of one of the other sides, say, side c:
Pc( xB - (( xB - xA) / 2 ), yB - (( yB - yA) / 2 ) )

Insert the values:
Pc( 6 - ((6 - (-2)) / 2 ), 0 - ((0 - 4 ) / 2) )
Pc( 2, 2).

Next we have to find the slope mc of c:
mc = ( yB - yA) / ( xB - xA)

Insert the values and you get:
mc = 1/2
The slope of the perpendicular mc' is the negative inverse of mc. Hence,
mc' = -2.

Now we have a point and a slope, so we can build the linear equation in point slope form, in order to find xM by equating it with a' := y = 4 :

c' := (y - yPc) = mc(x - xPc)

y - 2 = 2 (x - 2)
y = 2x - 2

a' = c'
4 = 2x - 2
x = 3.

We have determined M( 3, 4).

The equation of a circle is :
(x - xM)^2 + (y - yM)^2 = r^2

Let's insert, say, point C( 6, 8) (and M, of course):

(6 - 3)^2 + (8 - 4)^2 = r^2
9 + 16 = 25
r = 5.

The equation of your circle is then, finally:

( x - 3 )² + ( y - 4 )² = 25.