To find the center and radius of the circle given by the equation \(x^2 + y^2 + 8x - 6y - 11 = 0\), we will complete the square for both \(x\) and \(y\).
First, rearrange the equation:
\[ x^2 + 8x + y^2 - 6y = 11 \]
Next, we will complete the square for the \(x\) terms and the \(y\) terms.
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Completing the square for \(x^2 + 8x\):
- Take half of the coefficient of \(x\) (which is 8), square it: \((\frac{8}{2})^2 = 16\).
- Thus, \(x^2 + 8x\) can be rewritten as \((x + 4)^2 - 16\).
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Completing the square for \(y^2 - 6y\):
- Take half of the coefficient of \(y\) (which is -6), square it: \((\frac{-6}{2})^2 = 9\).
- Thus, \(y^2 - 6y\) can be rewritten as \((y - 3)^2 - 9\).
Now substitute these completed squares back into the equation:
\[ (x + 4)^2 - 16 + (y - 3)^2 - 9 = 11 \]
Combine like terms:
\[ (x + 4)^2 + (y - 3)^2 - 25 = 11 \]
Add 25 to both sides to isolate the completed squares:
\[ (x + 4)^2 + (y - 3)^2 = 36 \]
Now we can identify the center and radius of the circle from the equation \((x - h)^2 + (y - k)^2 = r^2\):
- The center \((h, k)\) is \((-4, 3)\).
- The radius \(r\) is \(\sqrt{36} = 6\).
Thus, the center and radius of the circle are:
Center: \((-4, 3)\), Radius: \(6\).
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