Two circular tanks have the edges x² + y² + 4x - 6y + 4 = 0 and x² + y² - 8x - 2y + 8 = 0, with…

Step 1. The radical axis. $S_1 - S_2 = 12x - 4y - 4 = 0$, which is $3x - y - 1 = 0$.

Step 2. The point. On the $x$-axis $y = 0$, so $3x = 1$ and $x = \dfrac{1}{3} \approx 0.333$.

Step 3. The length. $S_1\left(\tfrac{1}{3}, 0\right) = \dfrac{1}{9} + \dfrac{4}{3} + 4 = \dfrac{49}{9}$. The tangent length is $\sqrt{\dfrac{49}{9}} = \dfrac{7}{3} \approx 2.33$ metres. The second circle gives the same: $\dfrac{1}{9} - \dfrac{8}{3} + 8 = \dfrac{49}{9}$.