Chapter 9.2, Problem 29WE

To determine

To graph: The given inequality.

Answer to Problem 29WE

The center and the radius for the given inequality are $\left(-3,3\right)$ and $3$ .

Explanation of Solution

Given information: The given inequality is $x^2+y^2+6x-6y+9\le 0$ .

Formula used: ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation: Consider a circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ with center $\left(h,k\right)$ and radius $\left(r\right)$ .

Now, simplify the given inequality by making complete square,

  $\begin{array}{l}{x}^2+y^2+6x-6y+9\le 0\\ x^2++6x+9+y^2-6y+9\le 9\\ {\left(x+3\right)}^2+{\left(y-3\right)}^2\le 9\end{array}$

Since, $\left(3,3\right):0^2+0^2\le 9$

Hence, the circle should be a solid line and the inside region should be shaded.

Compare ${\left(x+3\right)}^2+{\left(y-3\right)}^2\le 9$ to ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Hence, the center $\left(h,k\right)=\left(-3,3\right)$ and $r=3$ .

Therefore, the required graph of ${\left(x+3\right)}^2+{\left(y-3\right)}^2\le 9$ is given below.