To find the center and radius of the circle defined by the equation \( x^2 + y^2 + 12x - 10y = -36 \), we need to rewrite this equation in standard form, which is \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center of the circle and \(r\) is the radius.
Starting from the given equation:
\[ x^2 + y^2 + 12x - 10y = -36 \]
- Rearranging to make it easier to complete the square:
\[ x^2 + 12x + y^2 - 10y = -36 \]
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Completing the square for the \(x\) terms:
- \( x^2 + 12x \) can be completed as follows:
- Take \(12\), halve it to get \(6\), and square it to get \(36\).
- So, \( x^2 + 12x = (x + 6)^2 - 36 \).
- \( x^2 + 12x \) can be completed as follows:
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Completing the square for the \(y\) terms:
- \( y^2 - 10y \) can be completed similarly:
- Take \(-10\), halve it to get \(-5\), and square it to get \(25\).
- So, \( y^2 - 10y = (y - 5)^2 - 25 \).
- \( y^2 - 10y \) can be completed similarly:
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Now substitute back into the equation:
\[ (x + 6)^2 - 36 + (y - 5)^2 - 25 = -36 \]
Simplifying the left side, we get:
\[ (x + 6)^2 + (y - 5)^2 - 61 = -36 \]
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Move \( -61 \) to the other side:
\[ (x + 6)^2 + (y - 5)^2 = 25 \]
Now, we can identify the center and the radius:
- The center \((h, k)\) is \((-6, 5)\).
- The radius \(r\) is \(\sqrt{25} = 5\).
Thus, the center is \((-6, 5)\) and the radius is \(5\).
The correct answer is:
center: (−6, 5), radius: 5.