Circle Equation: Find The Radius Explained Simply

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Hey guys! Today, we're diving into a classic math problem: finding the radius of a circle when we're given its equation. Specifically, we'll tackle this equation: $x2+y2+8x-6y+21=0$. Don't worry if it looks intimidating; we'll break it down step-by-step so it's super easy to understand. We'll explore the core concepts behind circle equations, the method of completing the square, and finally, how to pinpoint that radius. So, grab your thinking caps, and let's get started!

Understanding the Circle Equation

Before we jump into solving the problem, let's rewind and quickly go through the fundamental equation of a circle. This foundational understanding is key to solving pretty much any circle-related problem. The standard equation of a circle is given by: $(x - h)^2 + (y - k)^2 = r^2$ Where:

  • (h, k) represents the coordinates of the center of the circle. Think of this as the circle's anchor point on the graph.
  • r is the radius of the circle. This is the distance from the center to any point on the circle's edge.

Now, why is this important? Well, our given equation, $x2+y2+8x-6y+21=0$, looks a bit different, right? It's in a general form. Our mission is to transform it into this standard form. By doing so, we can easily identify the circle's center and, most importantly, its radius. This transformation involves a clever technique called "completing the square," which we'll explore in detail in the next section. So keep this standard form in mind, because that's our target!

Why is the standard form so important?

The standard form isn't just some random arrangement of terms; it's a powerful tool that unlocks key information about the circle at a glance. Think of it as a secret code that reveals the circle's center and size instantly.

  • Center Location: The values of 'h' and 'k' directly tell you where the circle is positioned on the coordinate plane. No need to guess or estimate – it's right there in the equation.
  • Radius Size: The value of 'r' gives you the circle's radius, which determines its overall size. A larger 'r' means a bigger circle, and a smaller 'r' means a smaller circle.

Essentially, the standard form provides a clear and concise snapshot of the circle's essential properties. It eliminates any ambiguity and makes it incredibly easy to visualize and analyze the circle. This is why we're so keen on converting our given equation into this form. It's like putting on a pair of glasses that allows us to see the circle clearly!

The Power of Completing the Square

Okay, guys, here comes the real magic: completing the square! This technique is our secret weapon for transforming the general form of the circle equation into the standard form we discussed earlier. If you've never encountered this before, don't sweat it; we'll walk through it together. Completing the square is essentially an algebraic trick that allows us to rewrite quadratic expressions (those with x2x^2 and y2y^2 terms) in a more convenient form. In our case, it helps us create those perfect squared terms, $(x - h)^2$ and $(y - k)^2$, that we need for the standard circle equation.

Let's break down the process step-by-step:

  1. Rearrange and Group: First, we'll rearrange our equation, $x2+y2+8x-6y+21=0$, and group the x-terms and y-terms together, moving the constant term to the right side of the equation:

    (x2+8x)+(y2βˆ’6y)=βˆ’21(x^2 + 8x) + (y^2 - 6y) = -21

    We've created some space within the parentheses because this is where the magic will happen!

  2. Complete the Square for x: Now, focus on the x-terms: $x^2 + 8x$. To complete the square, we take half of the coefficient of our x-term (which is 8), square it, and add it to both sides of the equation. Half of 8 is 4, and 4 squared is 16. So, we add 16 inside the first set of parentheses:

    (x2+8x+16)+(y2βˆ’6y)=βˆ’21+16(x^2 + 8x + 16) + (y^2 - 6y) = -21 + 16

    Notice that we also added 16 to the right side of the equation to maintain balance. The expression inside the first parentheses is now a perfect square trinomial, meaning it can be factored into the form $(x + a)^2$.

  3. Complete the Square for y: We repeat the same process for the y-terms: $y^2 - 6y$. Half of -6 is -3, and (-3) squared is 9. So, we add 9 inside the second set of parentheses and also to the right side:

    (x2+8x+16)+(y2βˆ’6y+9)=βˆ’21+16+9(x^2 + 8x + 16) + (y^2 - 6y + 9) = -21 + 16 + 9

    Again, the expression inside the second parentheses is now a perfect square trinomial.

  4. Factor and Simplify: Now, we factor the perfect square trinomials and simplify the right side of the equation:

    • (x^2 + 8x + 16)$ factors into $(x + 4)^2

    • (y^2 - 6y + 9)$ factors into $(y - 3)^2

    • -21 + 16 + 9$ simplifies to 4

    Our equation now looks like this:

    (x+4)2+(yβˆ’3)2=4(x + 4)^2 + (y - 3)^2 = 4

    Boom! We've done it! We've successfully transformed the equation into the standard form of a circle equation.

Why does completing the square work?

Completing the square might seem like a magical trick, but there's a solid algebraic reason why it works. It's based on the fundamental pattern of squaring a binomial:

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

Notice that the constant term ($b^2$) is the square of half the coefficient of the 'a' term (2b). This is the core idea behind completing the square. By adding the correct constant term, we force the expression to fit this perfect square pattern, allowing us to factor it easily. Think of it as adding the missing puzzle piece to complete the square!

Finding the Radius: The Final Step

Alright, guys, the hard work is done! We've successfully transformed the equation $x2+y2+8x-6y+21=0$ into its standard form: $(x + 4)^2 + (y - 3)^2 = 4$. Now, it's time for the grand finale: identifying the radius. Remember, the standard equation of a circle is: $(x - h)^2 + (y - k)^2 = r^2$ Where 'r' is the radius.

Comparing Our Equation

Let's compare our transformed equation with the standard form:

  • Our equation: $(x + 4)^2 + (y - 3)^2 = 4$
  • Standard form: $(x - h)^2 + (y - k)^2 = r^2$

Notice that the right side of our equation is 4. This corresponds to $r^2$ in the standard form. Therefore, we have: $r^2 = 4$ To find the radius 'r', we simply take the square root of both sides: $r = \sqrt{4} = 2$ So, the radius of the circle is 2 units!

Spotting the Center

While we were focused on the radius, let's also take a quick look at the center of the circle. Remember, the center is represented by (h, k) in the standard form. Looking at our equation, $(x + 4)^2 + (y - 3)^2 = 4$, we can identify:

  • h = -4 (notice the sign change because it's (x - h) in the standard form)
  • k = 3

So, the center of the circle is at the point (-4, 3). We now have a complete picture of the circle – its center and its radius!

Conclusion: Circle Equation Mastery

Awesome job, guys! You've successfully navigated the world of circle equations and conquered the challenge of finding the radius. We started with a seemingly complex equation, $x2+y2+8x-6y+21=0$, and, using the power of completing the square, transformed it into the standard form. This allowed us to easily identify the radius as 2 units and the center as (-4, 3). Remember, understanding the standard form of the circle equation and mastering the technique of completing the square are crucial skills in geometry and beyond. So, keep practicing, and you'll be a circle equation whiz in no time! You've got this!