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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 follow these steps:
1. Identify the Radius from the Diameter:
- The diameter given is 12 units. Therefore, the radius [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{12}{2} = 6 \][/tex]
2. Compare Given Equations with Circle Standard Form:
- We know that the standard form of the equation of a circle is [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], where [tex]\( (h, k) \)[/tex] is the center of the circle, and [tex]\( r \)[/tex] is the radius.
3. Check the Radius in Each Equation:
- The circle equation should have [tex]\( r^2 = 6^2 = 36 \)[/tex], so we need to see if the right-hand side of the equation is 36.
4. Check the Center on the [tex]\( y \)[/tex]-axis:
- For the center of the circle to lie on the [tex]\( y \)[/tex]-axis, the [tex]\( x \)[/tex]-coordinate [tex]\( h \)[/tex] must be 0. Therefore, we are looking for equations where the center [tex]\((h, k)\)[/tex] has [tex]\( h = 0 \)[/tex].
Let's analyze each equation:
- Equation 1: [tex]\( x^2 + (y-3)^2 = 36 \)[/tex]
- The center is [tex]\((0, 3)\)[/tex] and [tex]\( r^2 = 36 \)[/tex], satisfying both conditions.
- Equation 2: [tex]\( x^2 + (y-5)^2 = 6 \)[/tex]
- The center is [tex]\((0, 5)\)[/tex] but [tex]\( r^2 = 6 \)[/tex], which does not satisfy the radius condition.
- Equation 3: [tex]\( (x-4)^2 + y^2 = 36 \)[/tex]
- The center is [tex]\((4, 0)\)[/tex], which does not lie on the [tex]\( y \)[/tex]-axis.
- Equation 4: [tex]\( (x+6)^2 + y^2 = 144 \)[/tex]
- The center is [tex]\((-6, 0)\)[/tex] and [tex]\( r^2 = 144 \)[/tex], which does not satisfy the radius condition.
- Equation 5: [tex]\( x^2 + (y+8)^2 = 36 \)[/tex]
- The center is [tex]\((0, -8)\)[/tex] and [tex]\( r^2 = 36 \)[/tex], satisfying both conditions.
Based on these evaluations, the equations that represent circles with a diameter of 12 units and a center that lies 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]
Thus, the correct equations are:
[tex]\[ x^2+(y-3)^2=36 \quad \text{and} \quad x^2+(y+8)^2=36 \][/tex]
1. Identify the Radius from the Diameter:
- The diameter given is 12 units. Therefore, the radius [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{12}{2} = 6 \][/tex]
2. Compare Given Equations with Circle Standard Form:
- We know that the standard form of the equation of a circle is [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], where [tex]\( (h, k) \)[/tex] is the center of the circle, and [tex]\( r \)[/tex] is the radius.
3. Check the Radius in Each Equation:
- The circle equation should have [tex]\( r^2 = 6^2 = 36 \)[/tex], so we need to see if the right-hand side of the equation is 36.
4. Check the Center on the [tex]\( y \)[/tex]-axis:
- For the center of the circle to lie on the [tex]\( y \)[/tex]-axis, the [tex]\( x \)[/tex]-coordinate [tex]\( h \)[/tex] must be 0. Therefore, we are looking for equations where the center [tex]\((h, k)\)[/tex] has [tex]\( h = 0 \)[/tex].
Let's analyze each equation:
- Equation 1: [tex]\( x^2 + (y-3)^2 = 36 \)[/tex]
- The center is [tex]\((0, 3)\)[/tex] and [tex]\( r^2 = 36 \)[/tex], satisfying both conditions.
- Equation 2: [tex]\( x^2 + (y-5)^2 = 6 \)[/tex]
- The center is [tex]\((0, 5)\)[/tex] but [tex]\( r^2 = 6 \)[/tex], which does not satisfy the radius condition.
- Equation 3: [tex]\( (x-4)^2 + y^2 = 36 \)[/tex]
- The center is [tex]\((4, 0)\)[/tex], which does not lie on the [tex]\( y \)[/tex]-axis.
- Equation 4: [tex]\( (x+6)^2 + y^2 = 144 \)[/tex]
- The center is [tex]\((-6, 0)\)[/tex] and [tex]\( r^2 = 144 \)[/tex], which does not satisfy the radius condition.
- Equation 5: [tex]\( x^2 + (y+8)^2 = 36 \)[/tex]
- The center is [tex]\((0, -8)\)[/tex] and [tex]\( r^2 = 36 \)[/tex], satisfying both conditions.
Based on these evaluations, the equations that represent circles with a diameter of 12 units and a center that lies 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]
Thus, the correct equations are:
[tex]\[ x^2+(y-3)^2=36 \quad \text{and} \quad x^2+(y+8)^2=36 \][/tex]
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