Find the area of the region satisfying the inequality x^2 + y^2 \leq 4x + 6y+13.
\(x^2 + y^2 \leq 4x + 6y+13\\ x^2-4x + y^2-6y \leq 13\\ (x^2-4x )+ (y^2-6y) \leq 13\\ (x^2-4x+4 )+ (y^2-6y+9) \leq 13+4+9\\ (x-2 )^2+ (y-3)^2 \leq (\sqrt{26})^2\\\)
Oh you can find the area yourself. Just use the formula. :)
+150 
