How to prove that I've found the locus of the centers of the circles that pass through "these" two points?

Question

I have the next exercise:

Find the locus of the centers of the circles that pass through "these" two points

Here is the image with my solution drawing:

Sure, we can see that the locus I have to find is the mid perpendicular, but here is the point: to prove that this is the mid perpendicular I have to prove at first that all these centers of the circles are on the same straight line.

How to prove that ?

Answer 1

The problem can be defeated by Geometric concepts. As you indicated, it looks like the desired locus is simply all points on the perpendicular bisector of the two points. The following analysis verifies this.

A circle will pass through both points if and only if the circle has a center that is equidistant between the two points. This is because the circle will have a center with a radius, and the distance of each of the two points to the center of the circle must equal the radius. Therefore, the center of the circle must be equidistant between the two points.

So, the problem reduces to showing that any given 3rd point will be equidistant between the the two points if and only if the 3rd point is on the perpendicular bisector.

If the point is on the perpendicular bisector, you can form two triangles. Each triangle includes two of the pertinent vertices, with the 3rd vertex being the midpoint of the line segment connecting the first two points. It is easy to see that the two triangles are congruent. Therefore, any point on the perpendicular bisector must be equidistant from the two points.

For the analysis in the opposite direction, assume that a point is equidistant between the two points. Form two triangles, where each of the two triangles uses two of the pertinent vertices, and the 3rd vertex is the midpoint of the line segment connecting the first two points. Then, the two triangles will have a common side, and will also have sides that are equal because each side is (1/2) the length of the line segment between the two points.

Therefore, by side-side-side analysis, the corresponding two triangles are congruent. This implies that all of the angles in the two triangles are equal. This implies that each of the two triangles must be a right triangle. This implies that the center of the circle must be on the perpendicular bisector of the two points.

Answer 2

Actually I found the locus and proved it already more than 4 month ago, but that is my mistake, that I didn't write my proofs near the drawing. And solution is already there actually. I had to thought better yesterday.

So, here we go. To prove it we just simply need:

*At first, let's imagine that we have the same drawing as on the image I uploaded yesterday with the question, but we leave only circles and A and O points.

1. At first, let's name all the circles centers as X₁, X₂, etc. So the first thing we need is to join each X₁, X₂, etc. circles centers with both A and O points.
   *From here we will deal with the circle with X₁ center.

2. Here we know that X₁O X₁A are both radii, so the triangle AOX₁ is isosceles.

3. Lets lower a perpendicular to the AO line segment and name it X₁K. As AOX₁ is isosceles triangle, the perpendicular X₁K is also a bisector and a median. So here we proved that the center point of some circle named X₁ is on the mid perpendicular of the AO line segment. Proof-steps are the same for all the other circles center.

4. And according to the main property of angles deposition, we can't have several rays with the same degree measure of an angle(in our case, it's ∠90°), that are not matched to each other during the overlapping. That means that KX₁, KX₂ are on the same ray, what I needed to prove.

So here we proved that each circle center is on the mid perpendicular of the "AO" line segment, and this mid perpendicular is our locus.

Answer 3

Let $A=(a_1,a_2)$ and $B=(b_1,b_2)$ be the two points. A center $Q=(x,y)$ of a circle containing $A$ and $B$ must satisfy $$ \|A-Q\|^2 = \|B-Q\|^2, $$ which in coordinates is $$ (a_1-x)^2+(a_2-y)^2 = (b_1-x)^2 + (b_2-y)^2 $$ or $$ (a_1-x)^2 - (b_1-x)^2 = (b_2-y)^2 - (a_2-y)^2. $$ Using $s^2-t^2=(s-t)(s+t)$ on both sides leads to $$ (a_1+b_1-2x)(a_1-b_1) = (b_2+a_2-2y)(b_2-a_2) $$ which is the equation of a line. Moreover, the line passes through $(A+B)/2$. Can you continue from here?