Which equations represent circles that have a diameter of 12 units and a center that lies on the [tex]y[/tex]-axis? Select two options.

A. [tex]x^2+(y-3)^2=36[/tex]

B. [tex]x^2+(y-5)^2=6[/tex]

C. [tex](x-4)^2+y^2=36[/tex]

D. [tex](x+6)^2+y^2=144[/tex]

E. [tex]x^2+(y+8)^2=36[/tex]



Answer :

To determine which equations represent circles that have a diameter of 12 units and a center that lies on the y-axis, follow these steps:

1. Calculate the radius of the circle:
- Given the diameter is 12 units, the radius is half of the diameter.
- Radius = Diameter / 2 = 12 / 2 = 6 units.

2. Identify the general equation of a circle:
- The equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex].

3. Determine the radius squared:
- Radius squared = [tex]\(6^2 = 36\)[/tex].

4. Assess the centers of the circles based on the equations provided:
- Since the center lies on the y-axis, [tex]\(h = 0\)[/tex]. Thus, the equation takes the form [tex]\(x^2 + (y-k)^2 = 36\)[/tex].

5. Check each given equation and see if it matches our criteria:
- [tex]\(x^2 + (y-3)^2 = 36\)[/tex]
- This equation represents a circle with center (0, 3) and radius squared equals 36. It meets the criteria.
- [tex]\(x^2 + (y-5)^2 = 6\)[/tex]
- This equation represents a circle with center (0, 5) but radius squared is 6, not 36. It does not meet the criteria.
- [tex]\((x-4)^2 + y^2 = 36\)[/tex]
- This equation represents a circle with center (4, 0). Since 4 is not on the y-axis, it does not meet the criteria.
- [tex]\((x+6)^2 + y^2 = 144\)[/tex]
- This equation represents a circle with center (-6, 0) and radius squared is 144, not 36. It does not meet the criteria.
- [tex]\(x^2 + (y+8)^2 = 36\)[/tex]
- This equation represents a circle with center (0, -8) and radius squared equals 36. It meets the criteria.

6. Select the valid equations:
- [tex]\(x^2 + (y-3)^2 = 36\)[/tex]
- [tex]\(x^2 + (y+8)^2 = 36\)[/tex]

Therefore, the equations that represent circles with a diameter of 12 units and a center that lies on the y-axis are:
- [tex]\(x^2 + (y-3)^2=36\)[/tex]
- [tex]\(x^2 + (y+8)^2=36\)[/tex]