Circle Geometry

      This proof came about from a challenge question in which three points A, B and C were given, and the origin O, had to be determined.  The method used, used three simultaneous equations, using the general form of a circle (x + a)² + (y + b)² = r² .  The way I solved it however, was a geometrical method using cartesian formulae for midpoints, perpendicular lines, and points of intersection.  It relies on the fact that any three non-collinear points are concyclic.

Fig. 4 A, B, C lie on the circle with centre O

The above diagram illustrates the points A, B and C.  Joining A to B, B to C and A to C, gives the midpoints M, M' and M'' respectively.  Joining A, B and C to O gives r, r₁ and r₂ respectively



             r = r₁ = r₂  (equal radii)


             ∴ △AOB, △BOC and △AOC are isos. △s


             ∴ OM ⊥ AC, OM' ⊥ AB and OM'' ⊥ BC              (median ⊥ base in isos △s)

             Let OM, OM' and PM'' be l, l₁ and l₂ respectively

           O lies on the point where l = l₁ = l₂

           ∴ The point of intersection of the lines ⊥ to the midpoints of the sides of any △ prescribes 

the centre of a circle passing through its vertices



      In order to apply this to the situation described initially, first find two of the midpoints.  Find the two lines perpendicular to these through the two midpoints (using point-gradient formula.  Lastly find the point that these two lines intersect, this is the centre.  Note that only two lines are needed as the third will pass through the same point.  If the three points are collinear (ie. don't form a △ or a circle), then all three lines will be parallel.