Let X be the centre of C and r be the radius of C.

Imagine drawing straight lines from X to each of A, B and C, and extending these lines so they stop at the circumference of C.

It must be that r = AX + 1 = BX + 1 = CX+1, and so AX = BX = CX.

Therefore X lies on the perpendicular bisectors of AB, BC and CA (which cross at a single point).

The gradient of AB is (8-1)/(6 - ^- 1) = 1, and the midpoint of AB is (2.5, 4.5). Therefore the equation of the perpendicular bisector of AB is y - 4.5 = ^-1(x - 2.5), which simplifies to x + y = 7. (Equation 1)

The gradient of AC is (8 - 4)/(6 - ^-2) = 0.5, and the midpoint of AC is (2, 6). Therefore the equation of the perpendicular bisector of AC is y - 6 = ^-2(x - 2), which simplifies to 2x + y = 10. (Equation 2)

Solving these two equations simultaneously, by subtracting equation 1 from equation 2, we obtain x = 3 and y = 4. Therefore the coordinates of X are (3, 4).

We can also now calculate any of AX, BX or CX. AX looks most straightforward, so that's a good choice.

AX² = (6 - 3)²+ (8 - 4)² = 3² + 4² = 5², and so AX = 5.

This means that r = 5 + 1 = 6 units.

So the centre of C is (3, 4) and its radius is 6 units.

Challenge 387: Tangent Trio