To determine the center and radius of the circle given by the equation [tex]\( x^2 + y^2 + 6y = 7 \)[/tex], we need to convert the equation into the standard form of a circle's equation.
The standard form of a circle's equation is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2, \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
Here's a step-by-step process for converting the given equation into this form:
1. Start with the given equation:
[tex]\[ x^2 + y^2 + 6y = 7. \][/tex]
2. We need to complete the square for the term involving [tex]\(y\)[/tex]. Focus on the [tex]\(y\)[/tex] terms:
[tex]\[ y^2 + 6y. \][/tex]
3. To complete the square, add and subtract the square of half the coefficient of [tex]\(y\)[/tex], which is [tex]\(\left(\frac{6}{2}\right)^2 = 9\)[/tex]:
[tex]\[ y^2 + 6y + 9 - 9. \][/tex]
This gives:
[tex]\[ y^2 + 6y + 9 - 9 = (y + 3)^2 - 9. \][/tex]
4. Substitute this back into the original equation:
[tex]\[ x^2 + (y + 3)^2 - 9 = 7. \][/tex]
5. Move the [tex]\(-9\)[/tex] to the right side of the equation to form the standard circle equation:
[tex]\[ x^2 + (y + 3)^2 = 16. \][/tex]
Now the equation is in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
From the equation [tex]\( x^2 + (y + 3)^2 = 16 \)[/tex], we can see that:
- The center [tex]\((h, k)\)[/tex] is [tex]\((0, -3)\)[/tex].
- The radius [tex]\(r\)[/tex] is [tex]\(\sqrt{16} = 4\)[/tex].
Hence, the coordinates of the center are [tex]\((0, -3)\)[/tex] and the radius is 4.
Among the provided options, the correct one is:
(2) center [tex]\((0, -3)\)[/tex] and radius 4.
