+195 Find the area of the region satisfying the inequality $x^2 + y^2 \leq 4x + 6y+13$
Find the radius of the circle with equation $9x^2-18x+9y^2+36y+44=0.$
+79 1. First, subtract $4x$ and $6y$ from both sides to set up completing the square: $$x^2 -4x + y^2 -6y \leq 13.$$We add $((-4)/2)^2 = 4$ to complete the square in $x$ and $((-6)/2)^2 = 9$ to complete the square in $y$: $$x^2 -4x + 4 + y^2 -6y + 9 \leq 13+4+9,$$so $$(x-2)^2 + (y-3)^2 \leq 26.$$The graph of this inequality is a solid circle with center $(2,3)$ and radius $\sqrt{26}$, so the desired area is $\pi\left(\sqrt{26}\right)^2=\boxed{26\pi}$.
2. First, we factor out the constants of the squared terms to get $9(x^2-2x)+9(y^2+4y)=-44.$ To complete the square, we need to add $\left(\dfrac{2}{2}\right)^2=1$ after the $-2x$ and $\left(\dfrac{4}{2}\right)^2=4$ after the $4y,$ giving$$9(x-1)^2+9(y+2)^2=-44+9+36=1.$$Dividing the equation by $9$ gives$$(x-1)^2+(y+2)^2=\frac{1}{9},$$so the radius is $\sqrt{\frac{1}{9}}=\boxed{\frac{1}{3}}.$
Cphill emoji
😎😎😎
