Let the area enclosed by the solution curve y=y(x) of the differential equation (d y)/(d x)+(x-2)/(y-b)=2, b >0, y (5)=0 is 13 π. Let y=y(x)…

Let the area enclosed by the solution curve $\displaystyle y=y(x)$ of the differential equation $\displaystyle \frac{d y}{d x}+\frac{x-2}{y-b}=2, \mathrm{~b}>0, \mathrm{y}(5)=0$ is $\displaystyle 13 \pi$. Let $\displaystyle y=y(x)$ intersect the $\displaystyle y$-axis at the points $\displaystyle P$ and $\displaystyle Q$. If the tangents to $\displaystyle y=y(x)$ at $\displaystyle P$ and $\displaystyle Q$ meet at $\displaystyle T$, then the area of $\displaystyle \triangle P T Q$ is