Points where $AX^2 +BX^2 = AB^2$

Question

I'm trying to solve the following problem from Israel Gelfand's Trigonometry textbook:

Two points, $A$ and $B$, are given in the plane. Describe the set of points $X$ such that $AX^2 + BX^2 = AB^2$.

They give the answer "a circle with its center at the midpoint of $AB$", but I can't figure out why that is. Any help would be appreciated.

Answer 1

It looks like you want $AB$ to be the hypotenuse of a right triangle with legs $AX$ and $BX$. Consider that any triangle inscribed in a circle with the diameter being one of its sides is a right triangle. We're just looking at all of those.

Answer 2

This is a circle with a diameter $AB$ since $\angle AXB = 90^{\circ}$ for each $X$.

---

Or if we introduce coordinate system, say $A= (-1,0)$ and $B= (1,0)$ and $X = (x,y)$ we get:

$$ (x+1)^2+y^2+(x-1)^2+y^2 = 4\;\;\; \Longrightarrow\;\;\; x^2+y^2 = 1$$

Answer 3

Have you heard about the cosine rule?

If yes, than the locus of points $X$ with the given condition are the points such that $\triangle AXB$ is right-angled with $\angle AXB=90$.

Another property of the right-angled triangles is that the median from the right angle is equal to half the hypothenuse. Hence your locus of points is a circle with radius half the hypothenuse and center its midpoint

Answer 4

Recall that the median in a triangle satisfies $$m_a^2=\frac{b^2+c^2}{2}-\frac{a^2}{4}$$

Let $O$ be the midpoint of $AB$. Then, by this formula, we have $$XO^2=\frac{XA^2+XB^2}{2}-\frac{AB^2}{4}=\frac{AB^2}{4}$$

This is a fixed constant, therefore, your locus is a circle with centre $O$ and radius $\frac{AB}{2}$.