Question

Find the position of the centre of the circle circumscribing the triangle whose vertices are the points $(2,3)$, $(3,4)$, and $(6,8)$.

Answer 1

Let $(h,k)$ be the centre of the circle

$r=\sqrt{{(h-2)}^2+{(k-3)}^2}=\sqrt{{(h-3)}^2+{(k-4)}^2}=\sqrt{{(h-6)}^2+{(k-8)}^2}$

$\sqrt{{(h-2)}^2+{(k-3)}^2}=\sqrt{{(h-3)}^2+{(k-4)}^2}{(h-2)}^2+{(k-3)}^2={(h-3)}^2+{(k-4)}^2{h}^2+4-4h+k^2+9-6k=h^2+9-6h+k^2+16-8k2h+2k=12h+k=\mathrm{6........}\left(i\right)$

Also $\sqrt{{(h-3)}^2+{(k-4)}^2}=\sqrt{{(h-6)}^2+{(k-8)}^2}$

${(h-3)}^2+{(k-4)}^2={(h-6)}^2+{(k-8)}^2{h}^2+9+-6h+k^2+16-8k=h^2+36-12h+k^2+64-16k6h+8k=\mathrm{75.......}\left(ii\right)$

substituting $\left(i\right)$ in $\left(ii\right)$

$6(6-k)+8k=752k=75-36k=\frac{39}{2}=19\frac{1}{2}h=6-\frac{39}{2}=-\frac{27}{2}=-13\frac{1}{2}$

So the coordinates of centre $(-13\frac{1}{2},19\frac{1}{2})$