Two circles' centerline ⊥ common chord

Depends on: Perpendicular bisector test (equidistant from two endpoints ⇒ on the perpendicular bisector).

Statement

Let $\odot O_1(r_1)$ and $\odot O_2(r_2)$ be two intersecting circles, with common points $A$ and $B$. Then the line of centers $O_1O_2$ is the Perpendicular bisector ⇔ equidistant from endpoints of the common chord $AB$ — that is,

This welds together "two circles + line of centers + common chord" in one stroke: the line of centers is automatically both perpendicular to and bisects the common chord, with no extra conditions required.