jmc

First of all, we can translate everything downwards by $76$ and to the left by $14$. Then, note that a line passing through a given point intersecting a circle with a center as that given point will always cut the circle in half, so we can re-phrase the problem: Two circles, each of radius $3$, are drawn with centers at $(0,16)$, and $(5,8)$. A line passing through $(3,0)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line? Note that this is equivalent to finding a line such that the distance from $(0,16)$ to the line is the same as the distance from $(5,8)$ to the line. Let the line be $y-ax-b=0$. Then, we have that:$$\frac{|-5a+8-b|}{\sqrt{a^2+1}}=\frac{|16-b|}{\sqrt{a^2+1}}⟺|-5a+8-b|=|16-b|$$We can split this into two cases. Case 1: $16-b=-5a+8-b⟺a=-\frac{8}{5}$ In this case, the absolute value of the slope of the line won’t be an integer, and since this is an AIME problem, we know it’s not possible. Case 2: $b-16=-5a+8-b⟺2b+5a=24$ But we also know that it passes through the point $(3,0)$, so $-3a-b=0⟺b=-3a$. Plugging this in, we see that $2b+5a=24⟺a=-24$. $\boxed{24}$.