A circle has the same center as an ellipse and passes through the foci $F_1$ and $F_2$ of the ellipse, two curves intersect in $4$ points. A circle has the same center as an ellipse and passes through the foci $F_1$ and $F_2$ of the ellipse,such that the two curves intersect in $4$ points.Let $P$ be any one of their point of intersection.If the major axis of the ellipse is $17$ and the area of the triangle $PF_1F_2$ is $30$,then find the distance between the foci.

Let the center of the ellipse and the circle be $(0,0)$
We are given $2a=$length of major axis$=17$.
Let the coordinates of foci be $F_1(c,0)$ and $F_2(-c,0)$
We need to find $2c$.
Area of $PF_1F_2=\frac{1}{2}\times 2c\times$perpendicular distance between $P$ and the axis of the major axis of the ellipse.
I do not know how to solve it further.

We may suppose that 
$$\text{the ellipse$\ :\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a\gt b\gt 0$}$$
$$\text{the circle$\ :\ x^2+y^2=a^2-b^2$}$$
As you wrote, we have
$$2a=17\quad\Rightarrow \quad a=\frac{17}{2}$$
Since
$$\frac{a^2-b^2-y^2}{a^2}+\frac{y^2}{b^2}=1\quad\Rightarrow\quad |y|=\frac{b^2}{\sqrt{a^2-b^2}}$$
we have
$$30=\frac 12\times 2\sqrt{a^2-b^2}\times \frac{b^2}{\sqrt{a^2-b^2}}\quad\Rightarrow\quad b=\sqrt{30}.$$
Thus, the answer is
$$2\sqrt{a^2-b^2}=2\sqrt{\left(\frac{17}{2}\right)^2-30}=\color{red}{13}.$$
