Analyzing The Circle Equation $x^2 + Y^2 - 2x - 8 = 0$ To Find Radius And Center

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Hey guys! Today, we're diving into the world of circles and equations. We've got a fun problem on our hands: analyzing the circle equation x2+y2βˆ’2xβˆ’8=0x^2 + y^2 - 2x - 8 = 0. Our mission? To figure out some key properties of this circle, like its radius and the location of its center. We'll be focusing on determining whether certain statements about this circle are true or false. So, let’s roll up our sleeves and get started!

Understanding the Circle Equation

Before we jump into solving, let's quickly recap the standard form of a circle equation. This will be our guiding star in this problem. The standard form looks like this:

(xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2

Where:

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

Our given equation, x2+y2βˆ’2xβˆ’8=0x^2 + y^2 - 2x - 8 = 0, isn't quite in this form yet. It's more like a mixed-up puzzle! To solve this, we need to use a technique called completing the square. This will help us transform our equation into the neat and tidy standard form.

Completing the square is a method used to rewrite a quadratic expression into a perfect square trinomial, which can then be factored into the form (xβˆ’h)2(x - h)^2 or (yβˆ’k)2(y - k)^2. This process involves manipulating the equation by adding and subtracting specific constants. These constants are chosen carefully to create the perfect square trinomials. Think of it like rearranging the pieces of a puzzle to fit together perfectly. In our case, we'll apply this technique to both the xx terms and the yy terms in the equation. This will allow us to rewrite the original equation in the standard form of a circle equation, making it easier to identify the circle's center and radius. Let's dive into the steps and see how it works in practice. So, stick with me, and we'll transform this equation into something much more manageable!

Completing the Square: A Step-by-Step Guide

Okay, let's get our hands dirty and start completing the square. This might sound intimidating, but trust me, it's like following a recipe. We just need to take it one step at a time.

Here's our equation again: x2+y2βˆ’2xβˆ’8=0x^2 + y^2 - 2x - 8 = 0

Step 1: Group the x terms and y terms together.

This makes it easier to focus on each variable separately. We get:

(x2βˆ’2x)+y2=8(x^2 - 2x) + y^2 = 8

Notice how we also moved the constant term (-8) to the right side of the equation. This sets the stage for creating our perfect square trinomials.

Step 2: Complete the square for the x terms.

Remember, the magic formula for completing the square is adding (b2)2(\frac{b}{2})^2 to both sides of the equation, where 'b' is the coefficient of the x term. In our case, b = -2.

So, we calculate (βˆ’22)2=(βˆ’1)2=1(\frac{-2}{2})^2 = (-1)^2 = 1. We add this to both sides:

(x2βˆ’2x+1)+y2=8+1(x^2 - 2x + 1) + y^2 = 8 + 1

Step 3: Factor the perfect square trinomial.

The expression inside the parentheses, x2βˆ’2x+1x^2 - 2x + 1, is now a perfect square! It factors beautifully into:

(xβˆ’1)2+y2=9(x - 1)^2 + y^2 = 9

Step 4: Rewrite the equation in standard form.

To make it crystal clear, we can rewrite y2y^2 as (yβˆ’0)2(y - 0)^2. This gives us:

(xβˆ’1)2+(yβˆ’0)2=9(x - 1)^2 + (y - 0)^2 = 9

Voila! We've successfully transformed our equation into the standard form of a circle equation. It's like we've unlocked the secret code! Now, we can easily read off the circle's center and radius. Isn't that satisfying? In the next section, we'll do just that – identify the center and radius and see which statements about our circle are actually true.

Identifying the Center and Radius

Alright, we've done the hard work of completing the square, and our equation is now shining in its standard form glory:

(xβˆ’1)2+(yβˆ’0)2=9(x - 1)^2 + (y - 0)^2 = 9

Now comes the fun part: identifying the center and radius. Remember the standard form equation? (xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2. The values of h, k, and r are staring right at us!

  • The center of the circle is (h, k). Looking at our equation, we can see that h = 1 and k = 0. So, the center is (1, 0).
  • The radius squared, r2r^2, is equal to 9. To find the radius r, we simply take the square root of 9. So, r = √9 = 3.

Therefore:

  • The center of our circle is (1, 0).
  • The radius of our circle is 3 units.

See how easy that was? By converting to standard form, we've revealed the circle's key characteristics. Now, let's put this knowledge to the test. We're ready to tackle those statements and see which ones hold true for our circle. Get ready to put on your detective hats, guys!

Evaluating the Statements

Okay, we've got the center (1, 0) and the radius 3. Time to put on our thinking caps and evaluate the statements given in the problem. This is where we use our newfound knowledge to determine what's true and what's false. Let's break it down:

Statement 1: The radius of the circle is 3 units.

We just calculated the radius, and guess what? It's indeed 3 units! So, this statement is TRUE. We're off to a good start!

Statement 2: The center of the circle lies on the x-axis.

Think about it: a point lies on the x-axis if its y-coordinate is 0. Our center is (1, 0). The y-coordinate is 0! So, this statement is also TRUE. Two for two – we're on a roll!

Statement 3: The center of the circle lies on the y-axis.

For a point to lie on the y-axis, its x-coordinate must be 0. Our center is (1, 0). The x-coordinate is 1, not 0. So, this statement is FALSE. We can't win them all, right?

In conclusion:

The true statements are:

  • The radius of the circle is 3 units.
  • The center of the circle lies on the x-axis.

We've successfully analyzed the circle equation and determined the correct statements. Give yourselves a pat on the back! In the next section, we'll wrap things up and reflect on what we've learned.

Wrapping Up

Woo-hoo! We've reached the end of our circle equation adventure. Let's take a moment to wrap up what we've accomplished. We started with a circle equation that looked a bit intimidating: x2+y2βˆ’2xβˆ’8=0x^2 + y^2 - 2x - 8 = 0. But we didn't let that scare us!

We used the powerful technique of completing the square to transform the equation into its standard form: (xβˆ’1)2+(yβˆ’0)2=9(x - 1)^2 + (y - 0)^2 = 9. From there, it was smooth sailing. We easily identified the center of the circle as (1, 0) and the radius as 3 units.

Finally, we put our knowledge to the test and evaluated the given statements. We confidently declared that:

  • The radius of the circle is 3 units (TRUE).
  • The center of the circle lies on the x-axis (TRUE).
  • The center of the circle lies on the y-axis (FALSE).

This problem was a fantastic exercise in understanding circle equations and using algebraic techniques to uncover their hidden properties. Remember, guys, math isn't just about formulas; it's about problem-solving and critical thinking. We took a complex problem, broke it down into manageable steps, and emerged victorious! So, keep practicing, keep exploring, and keep having fun with math. Until next time!