To find the center and radius of the circle given by the equation \(x^2 + y^2 - 16x - 6y = -24\), we need to complete the square for both the \(x\) and \(y\) terms.
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Start with the equation: \[ x^2 - 16x + y^2 - 6y = -24 \]
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Complete the square for the \(x\) terms: \[ x^2 - 16x \quad \text{can be written as} \quad (x^2 - 16x + 64 - 64) = (x - 8)^2 - 64 \]
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Complete the square for the \(y\) terms: \[ y^2 - 6y \quad \text{can be written as} \quad (y^2 - 6y + 9 - 9) = (y - 3)^2 - 9 \]
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Substitute these back into the equation: \[ (x - 8)^2 - 64 + (y - 3)^2 - 9 = -24 \]
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Combine like terms: \[ (x - 8)^2 + (y - 3)^2 - 73 = -24 \] \[ (x - 8)^2 + (y - 3)^2 = 49 \]
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Now, we can see that this is in the form of the standard equation of a circle: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \((h, k)\) is the center and \(r\) is the radius.
From our final equation, we have:
- The center \((h, k) = (8, 3)\)
- The radius \(r = \sqrt{49} = 7\)
Thus, the center and radius of the circle are:
- Center = \((8, 3)\), Radius = 7
The correct response is: center=(8,3), radius=7