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The circumcenter of a nondegenerate triangle is the common intersection of its three perpendicular bisectors. It is equidistant from the three vertices and is the center of the triangle's circumcircle.

Expanding the squared Euclidean distances givesThus one may takeThe normal is parallel to $\overrightarrow{AB}$, so $L_{AB}$ is its perpendicular bisector. If $\mathbf{x}$ lies on both $L_{AB}$ and $L_{BC}$, then $|\mathbf{x}-\mathbf{a}|=|\mathbf{x}-\mathbf{b}|=|\mathbf{x}-\mathbf{c}|$, hence it also lies on $L_{CA}$. Geometrically, the three perpendicular bisectors of the noncollinear triangle concur at its circumcenter.

By the focal definition of an ellipse, the solution set is the ellipse with foci $a$ and $b$. Its centre is $(a+b)/2$, its major axis lies along the line through the foci, and its semiaxes areThe sketch is consequently a nondegenerate ellipse symmetric about both the focal line and its perpendicular bisector.