Question:
A circle has the same center as an ellipse and passes through the foci $F_1$ and $F_2$ of the ellipse.  The two curves intersect in 4 points.  Let $P$ be any point of intersection.  If the major axis of the ellipse is 15 and the area of triangle $PF_1 F_2$ is 26, compute the distance between the foci.

Answer:
Let $x = PF_1$ and $y = PF_2.$  Then $x + y = 15$ and $\frac{1}{2} xy = 26,$ so $xy = 52.$

[asy]
unitsize(0.5 cm);

path ell = xscale(5)*yscale(3)*Circle((0,0),1);
pair P = intersectionpoints(ell,Circle((0,0),4))[1];
pair[] F;

F[1] = (-4,0);
F[2] = (4,0);

draw(ell);
draw(Circle((0,0),4));
draw((-5,0)--(5,0),dashed);
draw(F[1]--P--F[2]);
draw(rightanglemark(F[1],P,F[2],15));
 
dot("$F_1$", F[1], SW);
dot("$F_2$", F[2], SE);
dot("$P$", P, NW);
[/asy]

Since $P$ lies on the circle with diameter $\overline{F_1 F_2},$ $\angle F_1 PF_2 = 90^\circ.$  Then by Pythagoras,
\[(F_1 F_2)^2 = x^2 + y^2.\]Squaring the equation $x + y = 15,$ we get $x^2 + 2xy + y^2 = 225.$  Then $x^2 + y^2 = 225 - 2xy = 225 - 2 \cdot 52 = 121,$ so $F_1 F_2 = \boxed{11}.$