A 180 Problem

We have $A = (-5.5, 5 ), B = ( -5.5, -5), C = (5.5, -5), D = (5.5, 5)$ .

If $P$ is a point on the locus, then by translating the image down 10 units, we obtain that $BPCP'$ is a cyclic quadrilateral. Observe that trivially, $ABBC$ and $DCCB$ are cyclic quadrilaterals, which means that $A , D$ lie on the locus. Hence, with the assumption that the locus is a hyperbola of the form $x^2 - y^2 = K$ , we obtain that $K = 5.5^2 - 5^2 = 5.25$ .

Now, let's determine the locus properly. Suppose that $P = (X,Y) , P' = (X, Y-10)$ is on the locus. We will use the fact that $BPCP'$ is a cyclic quadrilateral. The center of the circle can be determined by the intersection of perpendicular bisectors. Since the perpendicular bisector of $BC$ is $x = 0$ and the perpendicular bisector of $P P '$ is $y = Y - 5$ , we get that the center of the circle is $E = (0, Y - 5 )$ . The radius of the circle is $EB = (5.5)^2 +(Y-5 + 5)^2$ . Hence, the point $P$ satisfies

$(X - 0)^2 + (Y - (Y-5) ) ^2 = 5.5 ^ 2 + Y^2$

Hence, $X^2 - Y^2 = 5.5^2 - 5^2 = 5.25$ .

My solution hinges on the very convenient presentation of the hyperbola, whereas Calvin Lin gave a much more general approach. I'll still give it nonetheless:

The hyperbola crosses through the point $(\sqrt{K},0)$ based on the information given. At this point, $\Delta PAD \cong \Delta PBC$ .

Since $\angle APD$ corresponds to $\angle BPC$ , and their sum is $180^{\circ}$ , it follows that $\angle APD = \angle BPC = 90^{\circ}$

The locus of all points that make a right triangle off of some fixed segment $AD$ is a circle with diameter $AD$ . Hence, $(\sqrt{K},0)$ must lie on a circle with center $(0,5)$ and radius $r=5.5$ .

$(y-5)^{2}+x^{2}=5.5^{2}$ . Plugging in point $(\sqrt{K},0)$ gives $K=5.5^{2}-5^{2}=\boxed{5.25}$ .