Three Orthogonal Circles

Let us call the distance between the centres of the circles of radii $r_1$ and $r_2$ as $r_{12}$ , and similarly the other two distances as $r_{23}$ and $r_{31}$ . Then

$r_{12}=\sqrt {r_1^2+r_2^2},r_{23}=\sqrt {r_2^2+r_3^2}, r_{31}=\sqrt {r_3^2+r_1^2}$ .

Then $r_1r_2r_3=\sqrt {\dfrac {r_1^2r_2^2+r_2^2r_3^2+r_3^2r_1^2}{4}}$

Implies $\dfrac {1}{r_1^2}+\dfrac{1}{r_2^2}+\dfrac{1}{r_3^2}=\boxed 4$ .

On circles with radii $r_1$ and $r_2$ , the following red triangle can be drawn, which shares a side with the original blue triangle:

Since one vertex of the red triangle is at the intersection of the two orthogonal circles, it is a right triangle, and since two of the legs are the radii $r_1$ and $r_2$ , by the Pythagorean Theorem the hypotenuse (and the side of the original blue triangle) is $a = \sqrt{r_1^2 + r_2^2}$ .

By a similar argument, $b = \sqrt{r_1^2 + r_3^2}$ and $c = \sqrt{r_2^2 + r_3^2}$ .

By a version of Heron's formula, the area of the original blue triangle is $A = \frac{1}{4}\sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}$ , which after substituting $a = \sqrt{r_1^2 + r_2^2}$ , $b = \sqrt{r_1^2 + r_3^2}$ , and $c = \sqrt{r_2^2 + r_3^2}$ , becomes $A = \frac{1}{4}\sqrt{((r_1^2 + r_2^2) + (r_1^2 + r_3^2) + (r_2^2 + r_3^2))^2 - 2((r_1^2 + r_2^2)^2 + (r_1^2 + r_3^2)^2 + (r_2^2 + r_3^2)^2)}$ , and simplifies to $A = \frac{1}{2}\sqrt{r_1^2r_2^2 + r_1^2r_3^2 + r_2^2r_3^2}$

We are also given that $A = r_1r_2r_3$ , so $A = r_1r_2r_3 = \frac{1}{2}\sqrt{r_1^2r_2^2 + r_1^2r_3^2 + r_2^2r_3^2}$ , which can be rearranged to:

$\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} = 4$

Therefore, $k = \boxed{4}$ .