Answer :
To determine which equations represent circles that have a diameter of 12 units and a center that lies on the [tex]\( y \)[/tex]-axis, let's go through the steps needed to find the correct equations.
1. Find the radius of the circle:
The diameter of the circle is given as 12 units. The radius [tex]\( r \)[/tex] can be found using the relationship:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{12}{2} = 6 \text{ units} \][/tex]
2. Determine the equation for the circle:
The general 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]
Since the circle’s center lies on the [tex]\( y \)[/tex]-axis, the [tex]\( x \)[/tex]-coordinate of the center is zero. So the equation simplifies to:
[tex]\[ x^2 + (y - k)^2 = r^2 \][/tex]
3. Calculate the value of [tex]\( r^2 \)[/tex]:
Given [tex]\( r = 6 \)[/tex]:
[tex]\[ r^2 = 6^2 = 36 \][/tex]
So the equations we are looking for are in the form:
[tex]\[ x^2 + (y - k)^2 = 36 \][/tex]
where [tex]\( k \)[/tex] could be any real number indicating the [tex]\( y \)[/tex]-coordinate of the center.
4. Check the provided options:
Let's examine each option to see if they fit the criteria:
- [tex]\( x^2 + (y - 3)^2 = 36 \)[/tex]:
This equation is of the form [tex]\( x^2 + (y - k)^2 = 36 \)[/tex], where [tex]\( k = 3 \)[/tex]. It is a valid equation representing a circle with a radius of 6 units and centered at [tex]\((0, 3)\)[/tex].
- [tex]\( x^2 + (y - 5)^2 = 6 \)[/tex]:
This equation is of the form [tex]\( x^2 + (y - k)^2 = r^2 \)[/tex], but here [tex]\( r^2 = 6 \)[/tex], which means the radius [tex]\( r \)[/tex] would be [tex]\(\sqrt{6}\)[/tex], not 6. Hence, this equation does not fit our criteria.
- [tex]\( (x - 4)^2 + y^2 = 36 \)[/tex]:
This equation represents a circle with center at [tex]\((4, 0)\)[/tex] and radius 6. The center is not on the [tex]\( y \)[/tex]-axis. So, it does not fit our criteria.
- [tex]\( (x + 6)^2 + y^2 = 144 \)[/tex]:
This equation represents a circle with center at [tex]\((-6, 0)\)[/tex] and radius [tex]\(\sqrt{144} = 12\)[/tex], and the center is not on the [tex]\( y \)[/tex]-axis. So, it does not fit our criteria.
- [tex]\( x^2 + (y + 8)^2 = 36 \)[/tex]:
This equation is of the form [tex]\( x^2 + (y - k)^2 = 36 \)[/tex], where [tex]\( k = -8 \)[/tex]. It is a valid equation representing a circle with a radius of 6 units and centered at [tex]\((0, -8)\)[/tex].
Therefore, the two options that represent circles with a diameter of 12 units and centers on the [tex]\( y \)[/tex]-axis are:
1. [tex]\( x^2 + (y - 3)^2 = 36 \)[/tex]
2. [tex]\( x^2 + (y + 8)^2 = 36 \)[/tex]
1. Find the radius of the circle:
The diameter of the circle is given as 12 units. The radius [tex]\( r \)[/tex] can be found using the relationship:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{12}{2} = 6 \text{ units} \][/tex]
2. Determine the equation for the circle:
The general 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]
Since the circle’s center lies on the [tex]\( y \)[/tex]-axis, the [tex]\( x \)[/tex]-coordinate of the center is zero. So the equation simplifies to:
[tex]\[ x^2 + (y - k)^2 = r^2 \][/tex]
3. Calculate the value of [tex]\( r^2 \)[/tex]:
Given [tex]\( r = 6 \)[/tex]:
[tex]\[ r^2 = 6^2 = 36 \][/tex]
So the equations we are looking for are in the form:
[tex]\[ x^2 + (y - k)^2 = 36 \][/tex]
where [tex]\( k \)[/tex] could be any real number indicating the [tex]\( y \)[/tex]-coordinate of the center.
4. Check the provided options:
Let's examine each option to see if they fit the criteria:
- [tex]\( x^2 + (y - 3)^2 = 36 \)[/tex]:
This equation is of the form [tex]\( x^2 + (y - k)^2 = 36 \)[/tex], where [tex]\( k = 3 \)[/tex]. It is a valid equation representing a circle with a radius of 6 units and centered at [tex]\((0, 3)\)[/tex].
- [tex]\( x^2 + (y - 5)^2 = 6 \)[/tex]:
This equation is of the form [tex]\( x^2 + (y - k)^2 = r^2 \)[/tex], but here [tex]\( r^2 = 6 \)[/tex], which means the radius [tex]\( r \)[/tex] would be [tex]\(\sqrt{6}\)[/tex], not 6. Hence, this equation does not fit our criteria.
- [tex]\( (x - 4)^2 + y^2 = 36 \)[/tex]:
This equation represents a circle with center at [tex]\((4, 0)\)[/tex] and radius 6. The center is not on the [tex]\( y \)[/tex]-axis. So, it does not fit our criteria.
- [tex]\( (x + 6)^2 + y^2 = 144 \)[/tex]:
This equation represents a circle with center at [tex]\((-6, 0)\)[/tex] and radius [tex]\(\sqrt{144} = 12\)[/tex], and the center is not on the [tex]\( y \)[/tex]-axis. So, it does not fit our criteria.
- [tex]\( x^2 + (y + 8)^2 = 36 \)[/tex]:
This equation is of the form [tex]\( x^2 + (y - k)^2 = 36 \)[/tex], where [tex]\( k = -8 \)[/tex]. It is a valid equation representing a circle with a radius of 6 units and centered at [tex]\((0, -8)\)[/tex].
Therefore, the two options that represent circles with a diameter of 12 units and centers on the [tex]\( y \)[/tex]-axis are:
1. [tex]\( x^2 + (y - 3)^2 = 36 \)[/tex]
2. [tex]\( x^2 + (y + 8)^2 = 36 \)[/tex]
