Find Circle Center And Radius: X²+y²-6x+8y=0

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the cool world of mathematics, specifically tackling a classic problem involving circles. You know, those perfectly round shapes that pop up everywhere from pizza pies to the orbits of planets. Understanding their properties is super fundamental, and today we've got a challenge for you: Find the coordinates of the center of the circle and the length of its radius given the equation $x^2+y^2-6x+8y=0$. This might seem a bit daunting at first glance, especially if you're used to the simpler form $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius. But don't sweat it! We're going to break it down step-by-step, making it as easy as pie (pun intended!). By the end of this, you'll be a pro at converting this general form into the standard form, unlocking the secrets of the circle's center and radius. So, grab your thinking caps, maybe a calculator, and let's get this mathematical adventure started!

Unlocking the Secrets: Converting the General Equation

Alright, let's get down to business, guys. The general form of a circle's equation, $x^2+y^2-6x+8y=0$, looks a little different from the standard form we're used to, which is $(x-h)^2 + (y-k)^2 = r^2$. Our main mission here is to transform the given equation into this standard form. How do we do that, you ask? The magic trick is called completing the square. It sounds complicated, but it's actually a pretty neat algebraic technique. We'll be grouping the $x$ terms together, the $y$ terms together, and then moving the constant term (if there were one) to the other side of the equation. In our case, the constant term is zero, which simplifies things a bit. So, let's rearrange our equation first: grab all the $x$'s and $y$'s and put them on one side, leaving the $=0$ on the other. It looks like this: $(x^2 - 6x) + (y^2 + 8y) = 0$. See? We've just clustered the related terms. This is the crucial first step. Now, we need to make those parentheses perfect squares. For the $x$ terms, we have $x^2 - 6x$. To complete the square, we take the coefficient of the $x$ term (-6), divide it by 2 (which gives us -3), and then square that result ($(-3)^2 = 9$). We add this number inside the parenthesis for the $x$ terms. Similarly, for the $y$ terms, we have $y^2 + 8y$. The coefficient of the $y$ term is 8. Divide it by 2 (which is 4), and square it ($4^2 = 16$). We add this number inside the parenthesis for the $y$ terms. Crucially, whatever we add to one side of the equation, we must also add to the other side to maintain balance. Since we're adding 9 and 16 to the left side, we must also add them to the right side. So, our equation transforms into: $(x^2 - 6x + 9) + (y^2 + 8y + 16) = 0 + 9 + 16$. Now, the magic really happens! The expressions inside the parentheses are now perfect squares. $(x^2 - 6x + 9)$ factors into $(x-3)^2$, and $(y^2 + 8y + 16)$ factors into $(y+4)^2$. And on the right side, $0 + 9 + 16$ simply equals 25. So, our equation is now in the beautiful standard form: $(x-3)^2 + (y+4)^2 = 25$. Boom! We've successfully converted the general equation into the standard form. This is where we can easily spot the center and the radius. It's all about strategic grouping and that handy completing-the-square technique, guys. It’s a fundamental skill that opens up a whole new understanding of circles and other conic sections. Keep practicing this, and it'll become second nature!

Pinpointing the Center: Coordinates Revealed

Now that we've got our circle's equation in the pristine standard form, $(x-3)^2 + (y+4)^2 = 25$, it's time to reveal the coordinates of its center. Remember the standard form formula, $(x-h)^2 + (y-k)^2 = r^2$? This formula is our golden ticket because it directly tells us the location of the center and the length of the radius. Comparing our equation $(x-3)^2 + (y+4)^2 = 25$ to the standard form, we can identify the values of $h$ and $k$. For the $x$ part, we have $(x-3)^2$. Comparing this to $(x-h)^2$, it's clear that $h = 3$. Now, for the $y$ part, we have $(y+4)^2$. This looks a little different from $(y-k)^2$ because of the plus sign. But don't let that fool you, guys! We can rewrite $(y+4)^2$ as $(y - (-4))^2$. Now, when we compare $(y - (-4))^2$ to $(y-k)^2$, we can see that $k = -4$. So, the coordinates of the center of our circle are $(h, k)$, which is (3, -4). It's that straightforward once the equation is in the right format! Finding the center is all about recognizing the pattern. The number being subtracted from $x$ is the x-coordinate of the center, and the number being subtracted from $y$ (or the negative of the number being added to $y$) is the y-coordinate of the center. This method works for any circle equation that can be put into standard form. It's a core concept in coordinate geometry and understanding these transformations is key to mastering more complex problems. So, take a moment to appreciate how neat this is – a single equation holding all the information about the circle's position and size. Pretty cool, right?

Measuring the Radius: The Length Unveiled

Alright, we've successfully identified the center of our circle as (3, -4). Now, let's talk about the other crucial piece of information: the radius. Remember our standard form equation: $(x-h)^2 + (y-k)^2 = r^2$? On the right side of this equation, we have $r^2$, which represents the square of the radius. In our converted equation, $(x-3)^2 + (y+4)^2 = 25$, the number on the right side is 25. Therefore, we know that $r^2 = 25$. To find the actual length of the radius, $r$, we need to take the square root of $r^2$. So, $r = \sqrt{25}$. The square root of 25 is 5. Since the radius represents a length, it must be a positive value. Thus, the length of the radius is 5 units. So, to recap, our circle has its center at the coordinates (3, -4) and extends outwards with a radius of 5 units. This means any point on the edge of the circle is exactly 5 units away from the center (3, -4). This information is super valuable for graphing the circle, understanding its extent, or even calculating its circumference ($2\pi r$) or area ($\pi r^2$). The radius is a fundamental measure of a circle's size, and being able to extract it from the equation is a key skill. It's like getting the vital stats of the circle right from its algebraic description. Mastering these concepts will definitely make future math endeavors a whole lot smoother, guys. Keep up the great work!

Final Answer and Choices Explained

So, after all our hard work with completing the square and comparing it to the standard form, we've determined that the center of the circle described by the equation $x^2+y^2-6x+8y=0$ is (3, -4) and the length of its radius is 5 units. Now, let's look at the choices provided to see which one matches our findings:

A. (3,-4), 5 units: This option perfectly matches our calculated center and radius. B. (-3,4), 5 units: This option has the correct radius but the wrong coordinates for the center. Remember, we found $h=3$ and $k=-4$, not -3 and 4. C. (-3,4), 25 units: This option has incorrect coordinates for the center and also gives the square of the radius ($r^2$) instead of the radius ($r$). D. (3,-4), 25 units: This option has the correct coordinates for the center but incorrectly states the radius as 25 instead of its square root, 5.

Therefore, the correct answer is A. (3,-4), 5 units. It's always a good idea to double-check your work, especially when dealing with signs and square roots, to ensure you select the right option. Great job tackling this problem, mathematicians!