A circle has the same Centre as an ellipse, such that the two curves intersect in 4 points. Let ‘P’ be any point of the intersection. If the major axis of the ellipse is 17 and the area of the triangle PF₁F₂ is 30 then the distance between the foci is

As centre is same so $∆P{F}_1{F}_2$ will be right angled triangle at P because $F_1{F}_2$  is diameter of the circle.

$$\begin{gathered} Given area of ∆P{F}_1{F}_2=30 \\ \Rightarrow \frac{1}{2}\left(P{F}_1\right)\left(P{F}_2\right)=30 \\ \Rightarrow \left(P{F}_1\right)\left(P{F}_2\right)=60\to \left(1\right) \\ We know that P{F}_1+P{F}_2=2a=17\to \left(2\right) \\ Now {\left(F_1{F}_2\right)}^2={\left(P{F}_1\right)}^2+{\left(P{F}_2\right)}^2={(P{F}_1+P{F}_2)}^2-2\left(P{F}_1\right)\left(P{F}_2\right) \\ \Rightarrow F_1{F}_2=13 \end{gathered}$$

OR

Radius of the circle = ae
Eqⁿ of the circle is x²+y² = a²e²
‘P’ is $\left(\frac{a}{e}\sqrt{2e^2-1},\frac{a}{e}(1-e^2)\right)$

$\Rightarrow \frac{1}{2}\times 2ae\times \frac{a}{e}(1-e^2)=30$

$\Rightarrow a^2(1-e^2)=30\Rightarrow e=\frac{13}{17}$

$\Rightarrow 2ae=17\times \frac{13}{17}=13$