the center and radius for each equation of a circle. \( (x+5)^{2}+(y-6)^{2}=121 \) 2. \( (x-3)^{2}+(y+9)^{2}=36 \)

**Step 1: Recognize the General Form** The standard form for a circle is given by \[ (x - h)^2 + (y - k)^2 = r^2 \] where \((h, k)\) is the center and \(r\) is the radius. --- **Step 2: Analyze Each Equation** 1. **For the equation \((x+5)^2+(y-6)^2=121\):** - Rewrite the \(x\)-term: \((x+5)^2\) can be written as \((x-(-5))^2\). Hence, \(h = -5\). - The \(y\)-term is already in the form \((y-6)^2\) so, \(k = 6\). - The right-hand side of the equation is \(121\). Since \(r^2 = 121\), we have \(r = \sqrt{121} = 11\). **Center:** \((-5, 6)\) **Radius:** \(11\) --- 2. **For the equation \((x-3)^2+(y+9)^2=36\):** - The \(x\)-term \((x-3)^2\) shows that \(h = 3\). - Rewrite the \(y\)-term: \((y+9)^2\) can be written as \((y-(-9))^2\). Hence, \(k = -9\). - The right-hand side is \(36\). Since \(r^2 = 36\), we have \(r = \sqrt{36} = 6\). **Center:** \((3, -9)\) **Radius:** \(6\) --- **Final Answers:** 1. For \((x+5)^2+(y-6)^2=121\): - Center: \((-5, 6)\) - Radius: \(11\) 2. For \((x-3)^2+(y+9)^2=36\): - Center: \((3, -9)\) - Radius: \(6\)