To find the center and radius of the circle given by the equation \( x^2 + y^2 - 8x + 2y = 8 \), we can complete the square for both \( x \) and \( y \).
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Start with the equation: \[ x^2 + y^2 - 8x + 2y = 8 \]
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Rearrange the equation: \[ x^2 - 8x + y^2 + 2y = 8 \]
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Complete the square for \( x \):
- Take the coefficient of \( x \) (-8), halve it to get -4, and square it to get 16.
- Add and subtract 16: \[ (x^2 - 8x + 16) - 16 \]
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Complete the square for \( y \):
- Take the coefficient of \( y \) (2), halve it to get 1, and square it to get 1.
- Add and subtract 1: \[ (y^2 + 2y + 1) - 1 \]
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Substitute these back into the equation: \[ (x^2 - 8x + 16) + (y^2 + 2y + 1) = 8 + 16 + 1 \] Which simplifies to: \[ (x - 4)^2 + (y + 1)^2 = 25 \]
From this equation, we can see that it is in the standard circle form \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center, and \( r \) is the radius.
- The center \( (h, k) \) is \( (4, -1) \).
- The radius \( r \) is \( \sqrt{25} = 5 \).
Thus, the center and radius of the circle are:
- Center: \( (4, -1) \)
- Radius: \( 5 \)
The correct response is: center=(4,−1), radius=5