Circles With Diameter 12 And Center On Y-Axis Identifying Correct Equations

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Before we dive into solving the problem of finding the equations of circles with a specific diameter and center location, let's establish a solid foundation by understanding the standard form of a circle's equation. This will be our key tool in identifying the correct options. The general equation of a circle in the Cartesian coordinate system is given by:

(x - h)² + (y - k)² = r²

Where:

  • (h, k) represents the coordinates of the circle's center.
  • r represents the radius of the circle.

This equation is derived from the Pythagorean theorem and the definition of a circle as the set of all points equidistant from a central point. The distance between any point (x, y) on the circle and the center (h, k) is always equal to the radius r. Understanding this fundamental equation is crucial for analyzing and manipulating circle equations.

Key Components and Their Significance

  1. (h, k) - The Center: The center coordinates (h, k) directly determine the circle's position on the coordinate plane. A change in either h or k will shift the circle horizontally or vertically, respectively. When h = 0, the center lies on the y-axis, and when k = 0, the center lies on the x-axis. This is a critical point for solving the given problem.

  2. r - The Radius: The radius r dictates the size of the circle. It is the distance from the center to any point on the circle. The radius is squared in the equation (r²), so when given the equation, you need to take the square root of the constant term to find the radius. The diameter, which is twice the radius (d = 2r), is another way to express the size of the circle. In our problem, the diameter is given as 12 units, so we know the radius must be 6 units.

  3. (x - h)² and (y - k)² - The Variables: The variables x and y represent the coordinates of any point on the circle's circumference. These terms capture the relationship between the x and y coordinates as they vary along the circle's edge. The squares in these terms reflect the application of the Pythagorean theorem in defining the circle's geometry.

Applying the Standard Form

To effectively use the standard form, you need to be able to extract the center and radius from a given equation, or conversely, construct the equation given the center and radius. Let's look at some examples:

  • If the equation is (x - 2)² + (y + 3)² = 16, the center is (2, -3) and the radius is √16 = 4.
  • If the center is (-1, 4) and the radius is 5, the equation is (x + 1)² + (y - 4)² = 25.

Understanding these components and how they fit into the standard equation is vital for solving problems involving circles. Now, with this knowledge, we can approach the specific problem of identifying circles with a diameter of 12 units and a center on the y-axis.

Problem Setup: Diameter and Center on the y-axis

Now, let's translate the problem's conditions into mathematical requirements. We are looking for circle equations that satisfy two crucial criteria:

  1. Diameter of 12 units: As we discussed, the diameter (d) is twice the radius (r), so a diameter of 12 units means the radius must be 6 units. In the standard equation, this translates to r² = 6² = 36. Therefore, we are looking for equations where the constant term on the right side is 36.

  2. Center lies on the y-axis: For a circle's center (h, k) to lie on the y-axis, the x-coordinate (h) must be 0. This is because any point on the y-axis has an x-coordinate of 0. In the standard equation, this means the term (x - h)² becomes (x - 0)², which simplifies to x². This indicates that there should be no horizontal shift in the x-term of the equation.

With these two conditions clearly defined, we can now systematically evaluate the given options.

Evaluating the Options

Let's analyze each equation provided, keeping in mind our two criteria: a radius of 6 (r² = 36) and a center on the y-axis (h = 0). We'll go through each option, determine its center and radius, and see if it meets both conditions. This step-by-step approach will ensure we don't miss any valid solutions.

  1. x² + (y - 3)² = 36

    • Center: Comparing this to the standard form, we see that h = 0 and k = 3. Thus, the center is (0, 3), which lies on the y-axis.
    • Radius: The right side of the equation is 36, so r² = 36. Taking the square root, we get r = 6. This matches our required radius.
    • Conclusion: This equation meets both criteria. It represents a circle with a radius of 6 and a center on the y-axis. So, this is one of the options.

  2. x² + (y - 5)² = 6

    • Center: Again, h = 0 and k = 5, so the center is (0, 5), which lies on the y-axis.
    • Radius: The right side is 6, so r² = 6. This means r = √6, which is not equal to 6. This circle does not have the required radius.
    • Conclusion: This equation does not meet the diameter requirement. We can eliminate this option.

  3. (x - 4)² + y² = 36

    • Center: Here, h = 4 and k = 0, so the center is (4, 0). This point does not lie on the y-axis because the x-coordinate is not 0.
    • Radius: The right side is 36, so r² = 36, meaning r = 6. While the radius is correct, the center is not.
    • Conclusion: This equation fails the center requirement. We can rule out this option.

  4. (x + 6)² + y² = 144

    • Center: We have h = -6 and k = 0, so the center is (-6, 0). This does not lie on the y-axis.
    • Radius: The right side is 144, so r² = 144, which means r = 12. This radius is also incorrect (it should be 6).
    • Conclusion: This equation fails both the center and radius criteria. We can eliminate this option.

  5. x² + (y + 8)² = 36

    • Center: We have h = 0 and k = -8, so the center is (0, -8), which lies on the y-axis.
    • Radius: The right side is 36, so r² = 36, which means r = 6. This matches our required radius.
    • Conclusion: This equation meets both criteria: a center on the y-axis and a radius of 6. This is the second correct option.

Final Answer: The Correct Equations

After carefully evaluating each option, we have identified two equations that represent circles with a diameter of 12 units and a center on the y-axis:

  1. x² + (y - 3)² = 36
  2. x² + (y + 8)² = 36

These equations satisfy both conditions, making them the correct solutions to the problem. This methodical approach of breaking down the problem into smaller parts, understanding the underlying concepts, and systematically evaluating the options ensures accuracy and a clear understanding of the solution. In summary, by applying the standard equation of a circle and carefully considering the given conditions, we successfully identified the two correct options.