To solve this problem, we need to determine the possible equations for a circle with a diameter of 12 units, centered on the x-axis.
### Step-by-Step Solution:
1. Determine the Radius:
- The diameter of the circle is given as 12 units.
- The radius ([tex]\(r\)[/tex]) of a circle is half of its diameter.
[tex]\[
r = \frac{\text{Diameter}}{2} = \frac{12}{2} = 6 \text{ units}
\][/tex]
2. Identify Possible Centers on the x-axis:
- Since the circle’s center lies on the x-axis, the y-coordinate of the center is 0.
- The circle can be centered at any point [tex]\((a, 0)\)[/tex] on the x-axis.
3. Write the General Equation of the Circle:
- The standard form of the equation of a circle with center [tex]\((a, b)\)[/tex] and radius [tex]\(r\)[/tex] is:
[tex]\[
(x - a)^2 + (y - b)^2 = r^2
\][/tex]
- Plugging in the known values [tex]\(b = 0\)[/tex] and [tex]\(r = 6\)[/tex], the equation becomes:
[tex]\[
(x - a)^2 + y^2 = 6^2
\][/tex]
- Simplifying, we get:
[tex]\[
(x - a)^2 + y^2 = 36
\][/tex]
4. Check the Provided Options:
- Compare the provided equations to the general form [tex]\((x - a)^2 + y^2 = 36\)[/tex], considering various possible values for [tex]\(a\)[/tex].
- Option 1: [tex]\((x-12)^2 + y^2 = 12\)[/tex]
- Wrong radius squared [tex]\(12 \neq 36\)[/tex]. This doesn't match the required form.
- Option 2: [tex]\((x-6)^2 + y^2 = 36\)[/tex]
- Center at [tex]\((6, 0)\)[/tex] with radius 6. This is a valid equation.
- Option 3: [tex]\(x^2 + y^2 = 12\)[/tex]
- Neither proper radius squared nor consideration of center on the x-axis (not matching required distance). Incorrect form.
- Option 4: [tex]\(x^2 + y^2 = 144\)[/tex]
- Incorrect radius squared (144 instead of 36).
- Option 5: [tex]\((x+6)^2 + y^2 = 36\)[/tex]
- Center at [tex]\((-6, 0)\)[/tex] with radius 6. This is a valid equation.
- Option 6: [tex]\((x+12)^2 + y^2 = 144\)[/tex]
- Incorrect radius squared (144 instead of 36).
Therefore, the correct equations for the circles are:
[tex]\[
(x-6)^2 + y^2 = 36
\][/tex]
[tex]\[
(x+6)^2 + y^2 = 36
\][/tex]
