Circle Equation: Radius 5, Passes N(-2, 3), Center On X-Axis

Hey guys! Let's dive into a fun geometry problem: finding the equation of a circle. We've got a circle with a radius of 5, and it's passing through a specific point, N(-2, 3). But there's a catch – the center of this circle lies right on the x-axis. Sounds intriguing, right? Don't worry, we'll break it down step by step so it's super easy to understand.

Understanding the Basics of Circle Equations

Before we jump into the nitty-gritty, let's quickly refresh the basic equation of a circle. The standard form equation that represents any circle in the Cartesian plane is given by:

(x - h)² + (y - k)² = r²

Where:

• (x, y) represents any point on the circumference of the circle.
• (h, k) represents the coordinates of the center of the circle.
• r represents the radius of the circle.

This equation is derived directly from the Pythagorean theorem, considering the distances on the coordinate plane. Think of it this way: the equation is just a way of saying that the distance from any point on the circle to the center is always equal to the radius. So, if you know the center and the radius, you can define any circle! This foundational understanding is key to tackling more complex problems. So, make sure you're comfortable with this equation before we move on.

Now, why is this equation so useful? Well, it allows us to describe a circle using algebraic terms. This means we can use algebra to solve geometric problems, which is incredibly powerful. We can find things like the center of a circle, the radius, or even whether a point lies on a circle, all by manipulating this equation. This is the beauty of coordinate geometry: it bridges the gap between algebra and geometry, allowing us to solve problems using different perspectives. Keep this in mind as we move forward, and you'll see how smoothly everything comes together.

Key Elements We Need

To define our circle, we need two crucial pieces of information:

1. The Center (h, k): This is the heart of our circle. Since our circle's center lies on the x-axis, we know that its y-coordinate (k) is 0. So, our center will be in the form (h, 0). This is a fantastic piece of information because it simplifies our problem significantly. We've already knocked out one variable! Remember, in mathematics, any piece of information that reduces the number of unknowns is gold.
2. The Radius (r): We're given that the radius is 5. That's one less thing to worry about! Knowing the radius is like having a fixed measuring tape; it tells us how far every point on the circle is from the center. In our equation, this will be plugged in as r², so we'll be dealing with 5² = 25. Keep that number in mind!

So, with these two pieces, we're well on our way to defining our specific circle. We know the radius, and we know the form of the center. Now, all we need to do is find the exact value of 'h'. This is where the next piece of information comes in handy: the fact that the circle passes through the point N(-2, 3). Let's see how we can use this to our advantage!

Using the Point N(-2, 3) to Find the Center

Okay, here's where things get really interesting. We know the circle passes through the point N(-2, 3). This means that this point must satisfy the circle's equation. In other words, if we plug in x = -2 and y = 3 into our circle equation, it should hold true. This is a powerful concept: any point that lies on a curve (in this case, a circle) will always satisfy the equation of that curve. Think of it like a key fitting into a lock; the point perfectly fits the equation.

So, let's plug these values into our standard circle equation:

(x - h)² + (y - k)² = r²

becomes:

(-2 - h)² + (3 - 0)² = 5²

Notice how we've replaced x with -2, y with 3, and r with 5. Also, we've replaced k with 0 because we know the center lies on the x-axis. This substitution is a crucial step because it transforms our geometric problem into an algebraic one. We've now got an equation with only one unknown, 'h'. This is the beauty of using the coordinate system – it allows us to use algebraic techniques to solve geometric problems. The next step is to solve this equation for 'h', which will give us the x-coordinate of the circle's center.

Now, let’s simplify this equation and solve for 'h'. This is where our algebra skills come into play. Don't worry; we'll take it step by step so you can follow along easily. Remember, the goal is to isolate 'h' on one side of the equation. This will tell us the exact location of the circle's center on the x-axis, which is the final piece of the puzzle!

Solving for 'h'

Let's break down the equation we got after substituting the point N(-2, 3):

(-2 - h)² + (3 - 0)² = 5²

First, let's simplify the terms we can:

(-2 - h)² + 3² = 25

(-2 - h)² + 9 = 25

Now, let’s isolate the term with 'h' by subtracting 9 from both sides:

(-2 - h)² = 25 - 9

(-2 - h)² = 16

Next, we need to get rid of the square. We can do this by taking the square root of both sides. Remember, when you take the square root, you need to consider both positive and negative solutions:

√((-2 - h)²) = ±√(16)

-2 - h = ±4

Now we have two possible equations:

1. -2 - h = 4
2. -2 - h = -4

Let's solve each one separately:

For the first equation, -2 - h = 4, we add 2 to both sides:

-h = 6

h = -6

For the second equation, -2 - h = -4, we add 2 to both sides:

-h = -2

h = 2

So, we have two possible values for 'h': -6 and 2. This means there are actually two circles that satisfy the given conditions! This is a cool discovery, and it highlights the importance of considering all possibilities when solving math problems. Now that we have the possible values for 'h', we can determine the centers of our two circles.

Two Possible Circle Equations

We've found that there are two possible values for 'h': -6 and 2. This means we have two possible centers for our circle:

1. Center 1: (-6, 0)
2. Center 2: (2, 0)

Remember, we already knew that the y-coordinate of the center was 0 because the center lies on the x-axis. Now, we have the complete coordinates for both possible centers. This is a major step forward! With the centers and the radius, we can now write the equations for both circles. Are you excited? I am! This is where everything comes together and we see the final result of our hard work.

Now, let's plug these centers and the radius (r = 5) into the standard circle equation:

(x - h)² + (y - k)² = r²

For Center 1 (-6, 0):

(x - (-6))² + (y - 0)² = 5²

(x + 6)² + y² = 25

For Center 2 (2, 0):

(x - 2)² + (y - 0)² = 5²

(x - 2)² + y² = 25

So, we have two possible equations for the circle:

1. (x + 6)² + y² = 25
2. (x - 2)² + y² = 25

These are our final answers! We've successfully found the equations of two circles that meet all the given conditions. Both circles have a radius of 5, pass through the point N(-2, 3), and have their centers on the x-axis. This is a fantastic result, and it shows the power of combining geometry and algebra.

Conclusion

Wow, guys, we did it! We successfully found the equations of the circles that fit all the criteria. We started with a geometric problem and used our understanding of the circle equation and some algebraic techniques to solve it. Remember, the key steps were:

1. Understanding the standard circle equation.
2. Using the information about the center lying on the x-axis to simplify the problem.
3. Substituting the point N(-2, 3) into the equation.
4. Solving for 'h' to find the possible centers.
5. Plugging the centers and radius back into the standard equation to get the final circle equations.

This problem is a great example of how math can be like detective work. We were given some clues, and we used them to piece together the solution. And we didn't just find one solution; we found two! This highlights the importance of thinking critically and considering all possibilities.

So, next time you encounter a geometry problem, don't be intimidated. Break it down into smaller steps, use the tools you have, and remember that math can be a fun and rewarding puzzle to solve. Keep practicing, keep exploring, and you'll become a geometry whiz in no time! You've got this!