Circles: Find Center, Radius, And Graph Equations

by Kenji Nakamura 50 views

Hey guys! Today, we're diving deep into the fascinating world of circles and their equations. We'll be tackling the challenge of finding the center and radius of a circle from its equation and then graphing it on the Cartesian plane. Get ready to dust off your algebra and geometry skills – it's going to be a fun ride!

Understanding the Circle Equation

Before we jump into solving specific equations, let's make sure we're all on the same page about the standard form of a circle's equation. This is our key to unlocking the center and radius. The standard form is:

(x - h)² + (y - k)² = r²

Where:

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

Think of it like a secret code! Our mission is to transform the given equations into this form so we can easily read off the center and radius.

Completing the Square: Our Secret Weapon

Sometimes, the equations we're given aren't in the nice, neat standard form. They might look something like this:

x² + y² + Dx + Ey + F = 0

Don't worry, though! We have a powerful tool called completing the square that will help us convert these equations into standard form. Completing the square is an algebraic technique that allows us to rewrite quadratic expressions (expressions with x² or y²) as perfect squares. This is crucial for getting our equations into the (x - h)² and (y - k)² format.

The basic idea behind completing the square is to take a quadratic expression like x² + bx and add a constant term (b/2)² to it. This turns the expression into a perfect square trinomial, which can be factored as (x + b/2)². We'll apply this technique to both the x and y terms in our circle equations.

Think of it like this: we're taking a messy puzzle and rearranging the pieces to form a clear picture. Completing the square helps us see the hidden structure within the equation, revealing the circle's center and radius.

Problem A: x² + y² – 10x + 6y + 18 = 0

Let's kick things off with our first equation:

x² + y² – 10x + 6y + 18 = 0

Our goal is to rewrite this equation in the standard form (x - h)² + (y - k)² = r². To do this, we'll employ our trusty method of completing the square.

  1. Group the x and y terms:

    (x² – 10x) + (y² + 6y) + 18 = 0

    We've simply rearranged the terms to group the x terms together and the y terms together. The constant term (+18) is kept separate for now.

  2. Complete the square for the x terms:

    • Take half of the coefficient of the x term (-10), which is -5.
    • Square it: (-5)² = 25.
    • Add 25 inside the first parenthesis. Remember, to keep the equation balanced, we must also subtract 25 from the left side of the equation.

    (x² – 10x + 25) + (y² + 6y) + 18 - 25 = 0

    Now, the expression inside the first parenthesis is a perfect square trinomial.

  3. Complete the square for the y terms:

    • Take half of the coefficient of the y term (6), which is 3.
    • Square it: (3)² = 9.
    • Add 9 inside the second parenthesis. Again, to keep the equation balanced, we must also subtract 9 from the left side of the equation.

    (x² – 10x + 25) + (y² + 6y + 9) + 18 - 25 - 9 = 0

    Now, the expression inside the second parenthesis is also a perfect square trinomial.

  4. Factor the perfect square trinomials:

    (x - 5)² + (y + 3)² + 18 - 25 - 9 = 0

    We've factored the x terms as (x - 5)² and the y terms as (y + 3)². See how completing the square allows us to rewrite these quadratic expressions in a much simpler form?

  5. Simplify the constant terms:

    (x - 5)² + (y + 3)² - 16 = 0

    We've combined the constant terms: 18 - 25 - 9 = -16.

  6. Move the constant term to the right side of the equation:

    (x - 5)² + (y + 3)² = 16

    Finally, we've isolated the squared terms on the left side and moved the constant term to the right side. This is it – we've successfully transformed the equation into standard form!

  7. Identify the center and radius:

    Now that our equation is in the form (x - h)² + (y - k)² = r², we can easily identify the center and radius:

    • Center: (h, k) = (5, -3)
    • Radius: r = √16 = 4

    So, the circle has a center at (5, -3) and a radius of 4 units.

Graphing the Circle

Now for the fun part – graphing the circle! To graph the circle, follow these steps:

  1. Plot the center: Locate the point (5, -3) on the Cartesian plane and mark it.
  2. Plot points using the radius: From the center, measure out the radius (4 units) in all four directions (up, down, left, and right). Mark these points.
  3. Sketch the circle: Connect the points you plotted with a smooth curve to form the circle. Try your best to make it look circular!

Problem B: 4x² + 4y² + 24x + 16y + 27 = 0

Let's tackle our second equation:

4x² + 4y² + 24x + 16y + 27 = 0

This equation looks a bit more intimidating than the first one, but don't worry! We can handle it. The key difference here is the presence of the coefficient '4' in front of the x² and y² terms. This means we need to do a little extra work before we can complete the square.

  1. Divide the entire equation by 4:

    x² + y² + 6x + 4y + 27/4 = 0

    Dividing by 4 simplifies the equation and gets rid of the coefficients in front of the squared terms. This is essential for completing the square.

  2. Group the x and y terms:

    (x² + 6x) + (y² + 4y) + 27/4 = 0

    Just like before, we're grouping the x terms together and the y terms together.

  3. Complete the square for the x terms:

    • Take half of the coefficient of the x term (6), which is 3.
    • Square it: (3)² = 9.
    • Add 9 inside the first parenthesis and subtract 9 from the left side of the equation.

    (x² + 6x + 9) + (y² + 4y) + 27/4 - 9 = 0

  4. Complete the square for the y terms:

    • Take half of the coefficient of the y term (4), which is 2.
    • Square it: (2)² = 4.
    • Add 4 inside the second parenthesis and subtract 4 from the left side of the equation.

    (x² + 6x + 9) + (y² + 4y + 4) + 27/4 - 9 - 4 = 0

  5. Factor the perfect square trinomials:

    (x + 3)² + (y + 2)² + 27/4 - 9 - 4 = 0

    We've factored the x terms as (x + 3)² and the y terms as (y + 2)².

  6. Simplify the constant terms:

    (x + 3)² + (y + 2)² + 27/4 - 36/4 - 16/4 = 0

    (x + 3)² + (y + 2)² - 25/4 = 0

    We've found a common denominator (4) and combined the constant terms.

  7. Move the constant term to the right side of the equation:

    (x + 3)² + (y + 2)² = 25/4

    We've successfully transformed the equation into standard form!

  8. Identify the center and radius:

    • Center: (h, k) = (-3, -2)
    • Radius: r = √(25/4) = 5/2 = 2.5

    So, the circle has a center at (-3, -2) and a radius of 2.5 units.

Graphing the Circle

Let's graph this circle using the same steps as before:

  1. Plot the center: Locate the point (-3, -2) on the Cartesian plane and mark it.
  2. Plot points using the radius: From the center, measure out the radius (2.5 units) in all four directions and mark these points. Remember, 2.5 is halfway between 2 and 3!
  3. Sketch the circle: Connect the points with a smooth curve to form the circle. Nice job on your circle-drawing skills!

Conclusion

And there you have it! We've successfully navigated the world of circle equations, learned how to find the center and radius using the powerful technique of completing the square, and even graphed these circles on the Cartesian plane. You guys are awesome!

Remember, the key is to transform the equation into the standard form (x - h)² + (y - k)² = r². Once you've done that, the center and radius are practically staring you in the face. Keep practicing, and you'll become circle equation masters in no time!