By inspection, the centre of the circle is at (8,8), since the three given points form a right-triangle in which the hypothenus is the diameter (0,8),(16,8).
The radius is half the diameter, namely (16-0)/2=8
The standard form of a circle with centre at (xo,yo) and radius of R is
(x/xo)2 + (y/yo)2 = R2
Can you take it from here?
If you are looking for the general case of a circle passing through three distinct points P1(x1,y1), P2(x2,y2) and P3(x3,y3), it can be obtained by evaluating the following 4x4 determinant:
x2 + y2 x y 1
x12 + y1sup>2 x1 y1 1
x22 + y2sup>2 x2 y2 1
x32 + y3sup>2 x3 y3 1
Sorry, the determinant does not show up very well, and there was a typo:
x2 + y2 x y 1
x12 + y12 x1 y1 1
x22 + y22 x2 y2 1
x32 + y32 x3 y3 1
Please ignore the following format testing:
x2+ y2 x1 y1 1
x12+y12 x1 y1 1
x22+y22 x2 y2 1
x32+y32 x3 y3 1
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