A straight road is the line 3x + 4y - 10 = 0 on a plan whose lengths are in hundreds of metres…

Step 1. The family. The circle touches the line at $(2, 1)$, so it is $(x - 2)^2 + (y - 1)^2 + \lambda(3x + 4y - 10) = 0$.

Step 2. The number $\lambda$. Put in $(0, 3)$: $4 + 4 + \lambda(12 - 10) = 0$, so $\lambda = -4$.

Step 3. The circle. $x^2 - 4x + 4 + y^2 - 2y + 1 - 12x - 16y + 40 = 0$, which is $x^2 + y^2 - 16x - 18y + 45 = 0$. The centre is $(8, 9)$ and the radius is $\sqrt{64 + 81 - 45} = 10$.