When the circle leads to the square

Multiply the first equation by $2$ and add it to the second to obtain $17x^2+17y^2-68x-17=0$ .

This is the equation of a circle with center at $(2,0)$ and radius $r=\sqrt{5}$ , as we can see by doing the following transformations:

• Divide by $17$ : $x^2+y^2-4x-1=0$

• Build a perfect square with the terms that contain $x$ : $x^2-4x\textcolor{#D61F06}{+4}+y^2\textcolor{#D61F06}{-4}-1=0 \implies (x-2)^2+y^2=5$ .

The square we are looking for must be inscribed in this circle and its side is $l=\sqrt{2}r=\sqrt{10}$ . The perimeter is therefore $\boxed{4\sqrt{10}\approx12.65}$ .

This is a nice way of answering the question without having to find the four vertices of the square, which are $(0,-1), (1,2), (4,1)$ and $(3,-2)$ .

If you notice, both ellipses pass through the points $(1,2),(4,1),(3,-2),(0,-1)$ . This forms a square with side length $\sqrt{10}$ (because each square side is the hypotenuse of a triangle with sides $1$ and $3$ , and so using Pythagorean Theorem, we have $\sqrt{1^2+3^2}=\sqrt{10}$ ). So the perimeter is $4 \times \sqrt{10}=12.649$ .