Decoding Circle Equations Identifying Smallest And Largest Radii

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Hey guys! Let's dive into the fascinating world of circles and their equations. We've got a bunch of equations here, and our mission, should we choose to accept it, is to figure out which ones represent circles and, more importantly, which ones have the smallest and largest radii. Buckle up, because we're about to embark on a mathematical adventure!

Understanding the Standard Form of a Circle Equation

Before we jump into the nitty-gritty, let's quickly refresh our memory on the standard form of a circle equation. It's like the secret code to unlocking the circle's properties. The standard form looks like this:

(x - h)² + (y - k)² = r²

Where:

  • (h, k) is the center of the circle – think of it as the circle's home address.
  • r is the radius – the distance from the center to any point on the circle's edge.

Knowing this standard form is crucial because it allows us to easily identify the center and radius of a circle just by looking at its equation. Pretty neat, huh?

Transforming Equations into Standard Form

Now, here's the catch. The equations we're given might not always be in the nice, neat standard form. They might be hiding in a more general form, like this:

Ax² + Ay² + Bx + Cy + D = 0

Notice how the coefficients of x² and y² are the same (A) – that's a key indicator that we're dealing with a circle. But to find the center and radius, we need to transform this general form into the standard form. And how do we do that? By using a technique called completing the square.

Completing the square might sound intimidating, but it's actually a pretty straightforward process. It involves manipulating the equation to create perfect square trinomials for both the x and y terms. Let's break it down step by step:

  1. Group the x terms and y terms together: Rearrange the equation so that all the x terms are next to each other, and all the y terms are together.
  2. Move the constant term to the right side: Get that lone number hanging out on the left side and move it over to the right side of the equation.
  3. Complete the square for x: Take half of the coefficient of the x term (B/2A), square it ((B/2A)²), and add it to both sides of the equation. This will create a perfect square trinomial for the x terms.
  4. Complete the square for y: Do the same thing for the y terms. Take half of the coefficient of the y term (C/2A), square it ((C/2A)²), and add it to both sides of the equation. This will create a perfect square trinomial for the y terms.
  5. Factor the perfect square trinomials: The x terms and y terms should now be factorable into squared binomials, like (x - h)² and (y - k)².
  6. Simplify the right side: Combine the constants on the right side of the equation.

Voila! You've successfully transformed the equation into standard form. Now you can easily identify the center (h, k) and the radius (r) of the circle.

Analyzing Our Equations

Okay, enough theory. Let's get our hands dirty with some actual equations. We're given:

  1. x² + y² + 6x - 4y - 20 = 0
  2. 2x² + 2y² + 16x - 4y + 30 = 0

Our mission is to transform these equations into standard form, find their radii, and then determine which circle has the smallest radius and which has the largest.

Equation 1: x² + y² + 6x - 4y - 20 = 0

Let's tackle this one first. Follow along as we complete the square step by step.

  1. Group terms: (x² + 6x) + (y² - 4y) = 20
  2. Complete the square for x: Half of 6 is 3, and 3² is 9. Add 9 to both sides. (x² + 6x + 9) + (y² - 4y) = 20 + 9
  3. Complete the square for y: Half of -4 is -2, and (-2)² is 4. Add 4 to both sides. (x² + 6x + 9) + (y² - 4y + 4) = 20 + 9 + 4
  4. Factor: (x + 3)² + (y - 2)² = 33

Boom! We've got it in standard form. The center of this circle is (-3, 2), and the radius squared is 33. So, the radius is √33, which is approximately 5.74.

Equation 2: 2x² + 2y² + 16x - 4y + 30 = 0

Now let's tackle the second equation. Notice that the coefficients of x² and y² are both 2. That means we need to divide the entire equation by 2 to get it into the standard form.

  1. Divide by 2: x² + y² + 8x - 2y + 15 = 0
  2. Group terms: (x² + 8x) + (y² - 2y) = -15
  3. Complete the square for x: Half of 8 is 4, and 4² is 16. Add 16 to both sides. (x² + 8x + 16) + (y² - 2y) = -15 + 16
  4. Complete the square for y: Half of -2 is -1, and (-1)² is 1. Add 1 to both sides. (x² + 8x + 16) + (y² - 2y + 1) = -15 + 16 + 1
  5. Factor: (x + 4)² + (y - 1)² = 2

We did it again! The center of this circle is (-4, 1), and the radius squared is 2. So, the radius is √2, which is approximately 1.41.

Identifying the Smallest and Largest Radii

Alright, we've done the hard work. Now comes the fun part – comparing the radii.

  • Circle 1: Radius ≈ 5.74
  • Circle 2: Radius ≈ 1.41

It's clear that Circle 2 has the smallest radius, and Circle 1 has the largest radius.

Conclusion

So, there you have it! We've successfully decoded the equations of two circles, transformed them into standard form, calculated their radii, and identified the circles with the smallest and largest radii. We've learned the power of completing the square and the importance of the standard form of a circle equation. Go forth and conquer more mathematical challenges, my friends!

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