Hello everyone

I think it's time we drew this to a close. Here's what I got for the circumcentre, continuing from the point that I and **red_dog** had left at.

Solving the equations simultaneously, it's not difficult, but the arithmetic is tedious:

$\displaystyle (x-82)^2+(y-21)^2=(x-12)^2+(y-27)^2$

$\displaystyle \Rightarrow x^2-164x+82^2+y^2-42y+21^2=x^2-24x+12^2+y^2-54y+27^2$

$\displaystyle \Rightarrow 12y-140x+6724+441-144-729=0$

$\displaystyle \Rightarrow 12y-140x+6292=0$

$\displaystyle \Rightarrow 3y-35x+1573=0$ (1)

And:

$\displaystyle (x-12)^2+(y-27)^2=(x-74)^2+(y-43)^2$

$\displaystyle \Rightarrow x^2-24x+144+y^2-54y+729=x^2-148x+5476+y^2-86y+1849$

$\displaystyle \Rightarrow 32y+124x-6452=0$

$\displaystyle \Rightarrow 8y+31x-1613=0$ (2)

Multiply (1) by $\displaystyle 8$ and (2) by $\displaystyle 3$:

$\displaystyle 24y-280x+12584=0$

$\displaystyle 24y+93x-4839=0$

Subtract:$\displaystyle -373x+17423=0$

$\displaystyle \Rightarrow x = 46.71$ (to 2 d.p.)

Substitute into (1);$\displaystyle 3y-\frac{35\times17423}{373}+1573=0$

$\displaystyle \Rightarrow y = 20.62$ (to 2 d.p.)

So the circumcentre is $\displaystyle (46.71,20.62)$. Check that I haven't made any mistakes with the arithmetic.

Grandad