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

Ask by Curry Barber. in the United States
Mar 14,2025

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Answer

**Centers and Radii of the Circles:** 1. **First Circle:** - **Center:** \((-5, 6)\) - **Radius:** \(11\) 2. **Second Circle:** - **Center:** \((3, -9)\) - **Radius:** \(6\)

Solution

**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: \(6\)

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The center of the first circle is at the point (-5, 6) and the radius is 11, since the square root of 121 equals 11. This circle is centered in the upper left quadrant, providing a delightful spot for an imaginary picnic! For the second circle, the center is at the point (3, -9) and the radius is 6, as the square root of 36 equals 6. It's interesting to note that this circle is positioned in the lower right quadrant, perhaps setting the stage for a fun underwater adventure!

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