The Area enclosed between the ellipse $9x^2+4y^2-36x+8y+4=0$ and the line $3x+2y-10=0$ in the first quadrant. 

Given ellipse can be written as 

$\begin{array}{r}9(x-2{)}^2+4(y+1{)}^2=36\\ \Rightarrow \frac{(x-2{)}^2}{4}+\frac{(y+1{)}^2}{9}=1\end{array}$

Let $X = x - 2$ and $Y = y + 1.$ Then ellipse reduces to 

$\frac{X^2}{4}+\frac{Y^2}{9}=1$ and $3x + 2y - 10 = 0$ reduces to 

$3X+2Y-6=0$

Aria $=\frac{1}{4}\cdot \pi \cdot 2\cdot 3-\frac{1}{2}\cdot 2\cdot 3=\frac{3\pi }{2}-3$