Why is the locus of the centres of the circles passing through two points is the perpendicular bisector of the two points?

Question

Why is the locus of the centres of the circles passing through two points is the perpendicular bisector of the two points?

Answer 1

Take the centre $O$ of a circle through two points $AB$. Then $OA=OB$ because they are radii of the same circle. Hence $OAB$ is isosceles, and the altitude from $O$ to point $P$ on $AB$ defines two congruent triangles $OPA$ and $OPB$. So $OP$ is the perpendicular bisector of $AB$.

Any point $O$ on the perpendicular bisector likewise defines two congruent triangles so that $OA=OB$.

Answer 2

Let the two points be P & Q. so now, any circle passing through both P & Q,will have PQ as its chord.

Let there be two circles 1 & 2 with centres C & D respectively,satisfying above condition.

You'd have studied a theorem that perpendicular bisector of a chord of a circle passes through its centre.

Therefore, perpendicular bisector of chord PQ w.r.t circle 1,passes through C.

Similarly, perpendicular bisector of chord PQ w.r.t circle 2,passes through D.

Thus, the perpendicular bisector passes though both C & D, in other words, both centres C & D lie on the perpendicular bisector.

So you can say that line CD or in other words locus of centres of circles(passing though both P & Q) is the perpendicular bisector of line segment PQ.