To determine the coordinates of the center and the length of the radius of the circle given the equation [tex]\(x^2 + y^2 - 12y - 20.25 = 0\)[/tex], we'll rewrite this equation in the standard form of a circle's equation, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius.
1. Start with the given equation:
[tex]\[ x^2 + y^2 - 12y - 20.25 = 0 \][/tex]
2. To write this in the standard form, we need to complete the square for the [tex]\(y\)[/tex]-terms. First, we isolate the [tex]\(y\)[/tex]-terms and the constant:
[tex]\[ x^2 + (y^2 - 12y) = 20.25 \][/tex]
3. Next, complete the square for the [tex]\(y\)[/tex]-terms inside the parentheses. To do this, take half of the coefficient of [tex]\(y\)[/tex], square it, and add and subtract it inside the parentheses:
[tex]\[ y^2 - 12y \][/tex]
Half of [tex]\(-12\)[/tex] is [tex]\(-6\)[/tex], and [tex]\((-6)^2 = 36\)[/tex], so:
[tex]\[ y^2 - 12y + 36 - 36 \][/tex]
4. Now add 36 inside the parentheses and simultaneously add 36 to the right side of the equation to keep it balanced:
[tex]\[ x^2 + (y^2 - 12y + 36) - 36 = 20.25 \][/tex]
[tex]\[ x^2 + (y - 6)^2 - 36 = 20.25 \][/tex]
5. Simplify the equation by combining the constants on the right:
[tex]\[ x^2 + (y - 6)^2 = 20.25 + 36 \][/tex]
[tex]\[ x^2 + (y - 6)^2 = 56.25 \][/tex]
6. Now the equation is in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\(h = 0\)[/tex], [tex]\(k = 6\)[/tex], and [tex]\(r^2 = 56.25\)[/tex]. Thus:
[tex]\[ r = \sqrt{56.25} = 7.5 \][/tex]
Therefore, the coordinates of the center of the circle are [tex]\((0, 6)\)[/tex] and the radius is [tex]\(7.5\)[/tex].
Looking at the given options:
1) center [tex]\((0, 6)\)[/tex] and radius [tex]\(7.5\)[/tex]
2) center [tex]\((0, -6)\)[/tex] and radius [tex]\(7.5\)[/tex]
3) center [tex]\((0, 12)\)[/tex] and radius [tex]\(4.5\)[/tex]
4) center [tex]\((0, -12)\)[/tex] and radius [tex]\(4.5\)[/tex]
The correct answer is:
1) center [tex]\((0, 6)\)[/tex] and radius [tex]\(7.5\)[/tex]
