\[ \begin{array}{l}(x+2)^{2}+(y+6)^{2}=9 \\ \text { Part } 1 \text { of } 2 \\ \text { The center is }(\square, \square) \\ \text { Part } 2 \text { of } 2\end{array} \] The radius is \( r=\square \)

The given equation \((x+2)^{2}+(y+6)^{2}=9\) describes a circle in standard form, where the center can be derived from the expression inside the parentheses. Thus, the center of the circle is at \((-2, -6)\). As for the radius, the equation equals \(r^2\). Since 9 is the square of the radius, we find that the radius \(r\) is equal to 3 (because \(\sqrt{9} = 3\)). So, to summarize, the center is \((-2, -6)\) and the radius is \(3\).