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(C) Let $PF_1 = y$ and $PF_2 = x$. Since $P$ lies on the ellipse,by the definition of an ellipse,$x + y = 2a$,where $2a$ is the length of the major ax

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$13$

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(C) Let $PF_1 = y$ and $PF_2 = x$. Since $P$ lies on the ellipse,by the definition of an ellipse,$x + y = 2a$,where $2a$ is the length of the major axis. Given $2a = 17$,so $x + y = 17$.Since $P$ lies on the circle passing through $F_1$ and $F_2$ with center at the origin $O$,$PF_1$ and $PF_2$ are chords of the circle. From the geometry of the intersection,the angle $\angle F_1PF_2 = 90^\circ$ because the circle passes through the foci and the center is the origin (the diameter of the circle is the distance between the foci,$2ae$).Thus,$	riangle PF_1F_2$ is a right-angled triangle at $P$.The area of $	riangle PF_1F_2 = \frac{1}{2} 	imes x 	imes y = 30$,which implies $xy = 60$.The distance between the foci is the hypotenuse $F_1F_2 = \sqrt{x^2 + y^2}$.Using the identity $x^2 + y^2 = (x + y)^2 - 2xy$,we get:$F_1F_2^2 = (17)^2 - 2(60) = 289 - 120 = 169$.Therefore,$F_1F_2 = \sqrt{169} = 13$.

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