Dear Apurva
Although it's easier just to find the equation of the perpendicular  bisectors of two sides (mid point formula, gradient formula inverted  with sign changed, then use point gradient formula), and solve these  simultaneous equations.  That's what  did using points (x1, y1), (x2,  y2), (x3, y3).
 Calculate the determinants 
 D = | x1 .. x2 .. x3 |
 ,.,,, | y1 .. y2 .. y3 |
 ..... | 1 ..... 1 .... 1 |
 
 D1 = same as above, except that the x values are squared
 D2 = same as above, except that the y values are squared
 Then the circumcentre is given by
 x = [D1 - (y1 - y2)(y2 - y3)(y3 - y1)] / 2D
 y = [D2 - (x1 - x2)(x2 - x3)(x3 - x1)] / 2D
 
 For the orthocentre, it's not very hard to find the equation of the line  through A perpendicular to BC, the line through B perpendicular to AC,  and solve these simultaneous equations.  I haven't yet worked out a  formula for the result of that.  Will edit this if I do.  Email me if  you'd like an example.
 
 OK, I have a formula for the orthocentre.  Calculate the determinants
 D3 = | x2x3 .. x3x1 .. x1x2 |
 ....... | .. y1 ..... y2 ........y3 |
 ....... | ..1 ......... 1 ........ 1 .|
 D4 the same but with y2y3 etc in the top row.  Then
 x = -(D3 + D4)/D
 
 Now calculate D5 and D6, the same as these but with x replacing y in the middle row.  Then
 y = (D5 + D6)/D
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
 
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