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For the equation x² + y² - 2x - 6y - 6 = 0, do the following. Find the center (h,k) and radius r of the circle.

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Final answer:

The center of the circle is at (1, 3), and the radius is 4. We find this by completing the square to rewrite the given equation in the form (x - h)² + (y - k)² = r², revealing the center and radius.

Step-by-step explanation:

To find the center (h,k) and radius r of the circle described by the equation x² + y² - 2x - 6y - 6 = 0, we need to rewrite the equation in the standard form of a circle's equation, which is (x - h)² + (y - k)² = r². This involves completing the square for both the x and y terms.

First, we group the x and y terms and move the constant to the right side:
x² - 2x + y² - 6y = 6

Next, we add the squares of half the coefficients of x and y to each side to complete the square:
x² - 2x + 1 + y² - 6y + 9 = 6 + 1 + 9
This simplifies to (x - 1)² + (y - 3)² = 16

The center of the circle is at (h, k) = (1, 3), and the radius is r = √16 = 4.

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