Three posts stand at (1, 2), (5, 2) and (3, 6) on a field, with lengths in metres…

Step 1. The first bisector. The segment from $(1, 2)$ to $(5, 2)$ has midpoint $(3, 2)$ and is horizontal, so its perpendicular bisector is $x = 3$. The centre is $(3, k)$.

Step 2. Equal distances. The distance to $(1, 2)$ equals the distance to $(3, 6)$: $4 + (k - 2)^2 = (k - 6)^2$. So $4 + k^2 - 4k + 4 = k^2 - 12k + 36$, and $8k = 28$, giving $k = 3.5$.

Step 3. The radius. $r^2 = 4 + 1.5^2 = 6.25$, so $r = 2.5$ metres.

Step 4. The constant. $g = -3$, $f = -3.5$, so $c = g^2 + f^2 - r^2 = 9 + 12.25 - 6.25 = 15$. The equation is $x^2 + y^2 - 6x - 7y + 15 = 0$.