Circumcircle

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Suppose there exists a circle with equation (x - m)2 + (y - n)2 = r2 which passes through the three non-collinear points A(a , b), B(c , d) and C(e , f).

It is easy to derive the equation of normal to the circle at A(a , b):
(b - n)x - (a - m)y + (an - bm) = 0

Similarly, equation of normal to the circle at B(c , d):
(d - n)x - (c - m)y + (cn - dm) = 0

equation of normal to the circle at C(e , f):
(f - n)x - (e - m)y + (en - fm) = 0

For the above three equations, put x = m, y = n, we know that the three normals passes through the centre of the circle (m , n). With the perpendicular distance between the tangent to the circle and the centre of the circle = r

So, we prove there exist such circle that passes through the three non-collinear points.


Now, suppose there exists another circle with centre H(p , q) which passes through A, B and C.
Provided that p =/= m and q =/= n

However, if this is so, then the distance between H and the three points must not equal. That is, we cannot draw a circle such that it passes through the three points.

Contradiction occurs.

Therefore, H must equal to G.
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Hence, for the three non-collinear points on a Cartesian coordinates, there exists a UNIQUE circle with its circumference passing through all of them.

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