Let's analyze the given circle equation step by step:
1. Given equation:
[tex]\[
x^2 + y^2 + 4x - 6y - 36 = 0
\][/tex]
2. Move the constant term to the right side:
[tex]\[
x^2 + y^2 + 4x - 6y = 36
\][/tex]
3. Complete the square for the [tex]\(x\)[/tex] terms:
We need to complete the square for [tex]\(x\)[/tex]. The [tex]\(x\)[/tex] terms are [tex]\(x^2 + 4x\)[/tex].
- Add and subtract [tex]\(\frac{(4)}{2}^2 = 4\)[/tex]:
[tex]\[
x^2 + 4x = (x + 2)^2 - 4
\][/tex]
4. Complete the square for the [tex]\(y\)[/tex] terms:
Next, we complete the square for the [tex]\(y\)[/tex] terms. The [tex]\(y\)[/tex] terms are [tex]\(y^2 - 6y\)[/tex].
- Add and subtract [tex]\(\frac{(-6)}{2}^2 = 9\)[/tex]:
[tex]\[
y^2 - 6y = (y - 3)^2 - 9
\][/tex]
5. Rewrite the equation using the completed squares:
Substitute the expressions back into the original equation:
[tex]\[
(x + 2)^2 - 4 + (y - 3)^2 - 9 = 36
\][/tex]
Simplify:
[tex]\[
(x + 2)^2 + (y - 3)^2 - 13 = 36
\][/tex]
Move [tex]\(-13\)[/tex] to the other side:
[tex]\[
(x + 2)^2 + (y - 3)^2 = 49
\][/tex]
6. Identify the center and the radius:
The standard form of a circle's equation is:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
From [tex]\((x + 2)^2 + (y - 3)^2 = 49\)[/tex], we see:
[tex]\[
h = -2, \quad k = 3, \quad r^2 = 49 \quad \Rightarrow \quad r = \sqrt{49} = 7
\][/tex]
Thus, the center of the circle is [tex]\((-2, 3)\)[/tex] and the radius is [tex]\(7\)[/tex] units.
Now, let's evaluate the given statements:
1. To begin converting the equation to standard form, subtract 36 from both sides.
- True. We moved the constant term (-36) to the right side by adding 36 to both sides.
2. To complete the square for the [tex]\(x\)[/tex] terms, add 4 to both sides.
- True. For the [tex]\(x\)[/tex] terms, completing the square required adding and subtracting [tex]\(4\)[/tex].
3. The center of the circle is at [tex]\((-2,3)\)[/tex].
- True. We found the center to be [tex]\((-2, 3)\)[/tex].
4. The center of the circle is at [tex]\((4,-6)\)[/tex].
- False. The actual center is [tex]\((-2, 3)\)[/tex], not [tex]\((4, -6)\)[/tex].
5. The radius of the circle is 6 units.
- False. The radius is [tex]\(\sqrt{49} = 7\)[/tex] units, not 6.
6. The radius of the circle is 49 units.
- False. The radius squared ([tex]\(r^2\)[/tex]) is 49, but the radius ([tex]\(r\)[/tex]) itself is [tex]\(7\)[/tex] units.
Summary of True and False Statements:
- True Statements:
1. To begin converting the equation to standard form, subtract 36 from both sides.
2. To complete the square for the [tex]\(x\)[/tex] terms, add 4 to both sides.
3. The center of the circle is at [tex]\((-2, 3)\)[/tex].
- False Statements:
1. The center of the circle is at [tex]\((4, -6)\)[/tex].
2. The radius of the circle is 6 units.
3. The radius of the circle is 49 units.
This is the detailed step-by-step solution and verification of each statement.
