Example 3 Given the circle $x^{2}+y^{2}=r^{2}$ passes through the two foci $F_{1}(-c, 0)$, $F_{2}(c, 0)$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and the two curves have four intersection points, one of which is $P$. If the area of $\triangle F_{1} P F_{2}$ is 26, and the length of the major axis of the ellipse is 15, find the value of $a+b+c$.
(2000 "Hope Cup" Competition Question)
From the given, we know $\angle F_{1} P F_{2}=90^{\circ}$. By property 5 (II), we know $\left|P F_{1}\right| \cdot\left|P F_{2}\right|=\frac{b^{2}}{\cos ^{2} 45^{\circ}}=2 b^{2}$. Also, $\left|P F_{1}\right| \cdot\left|P F_{2}\right|=\frac{2 S_{\triangle F_{1} P F_{2}}}{\sin 90^{\circ}}=52$, so $b=\sqrt{26}$. Given $a=\frac{15}{2}$, then $c=\sqrt{a^{2}-b^{2}}=\frac{11}{2}$. Therefore, $a+b+c=13+\sqrt{26}$ is the answer.