A circular running track must pass through three marker posts at A(0, 0), B(10, 0) and C(0, 24). Where is the centre of the track?

The centre $(x,y)$ is equidistant from all three posts. From A and B: $x^2+y^2=(x-10{)}^2+y^2$ → $0=-20x+100$ → $x=5$. From A and C: $x^2+y^2=x^2+(y-24{)}^2$ → $0=-48y+576$ → $y=12$. Centre = (5, 12), and each distance is $\sqrt{25+144}=13$. The point $(\frac{10}{3},8)$ is the CENTROID (the average of the three posts) — a different centre altogether, and not equidistant from them; (5, 0) is only the midpoint of AB.