Finding The Radius Of A Circle From Its Equation

Hey there, math enthusiasts! Today, we're diving into the fascinating world of circles and their equations. We're going to tackle a specific problem, which involves finding the radius of a circle given its equation: $x^2 + y^2 - 10x + 6y + 18 = 0$. Sounds intimidating? Don't worry; we'll break it down step by step, making it super easy to understand. So, grab your thinking caps, and let's get started!

Decoding the Circle Equation The General Form

Before we jump into the problem, let's understand the basics. The general equation of a circle is given by:

$x^2 + y^2 + 2gx + 2fy + c = 0$

Where:

• (-g, -f) represents the center of the circle.
• $r = \sqrt{g^2 + f^2 - c}$ represents the radius of the circle.

This equation might look a bit complex at first, but it's essentially a blueprint that tells us everything we need to know about a circle its center and its size (radius).

Why is this equation so important? Well, it provides a concise way to describe a circle mathematically. Instead of drawing a circle on a graph, we can simply write down its equation. This is incredibly useful in various fields, from computer graphics to physics.

Key takeaway: The general form of a circle's equation is our starting point for finding the center and radius. Memorize it; it's your new best friend in circle-related problems!

Transforming to Standard Form Completing the Square

Our given equation, $x^2 + y^2 - 10x + 6y + 18 = 0$, looks similar to the general form but isn't quite there yet. To make things easier, we need to transform it into the standard form of a circle's equation:

$(x - h)^2 + (y - k)^2 = r^2$

Where:

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

This form is super helpful because it directly reveals the center and radius. So, how do we get there? The magic lies in a technique called completing the square.

Completing the square is a clever algebraic method that allows us to rewrite quadratic expressions (expressions involving terms like $x^2$ and $y^2$) in a more convenient form. It involves adding and subtracting specific constants to create perfect square trinomials (expressions that can be factored into the form $(x + a)^2$ or $(y + b)^2$).

Let's apply this to our equation:

1. Group the x and y terms:

   $(x^2 - 10x) + (y^2 + 6y) + 18 = 0$

2. Complete the square for the x terms: Take half of the coefficient of the x term (-10), square it ((-5)^2 = 25), and add and subtract it within the x group:

   $(x^2 - 10x + 25 - 25) + (y^2 + 6y) + 18 = 0$

3. Complete the square for the y terms: Take half of the coefficient of the y term (6), square it ((3)^2 = 9), and add and subtract it within the y group:

   $(x^2 - 10x + 25 - 25) + (y^2 + 6y + 9 - 9) + 18 = 0$

4. Rewrite as squared terms: Now, we can rewrite the expressions in parentheses as perfect squares:

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

5. Simplify: Combine the constant terms:

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

   $(x - 5)^2 + (y + 3)^2 - 16 = 0$

6. Move the constant to the right side: Add 16 to both sides:

   $(x - 5)^2 + (y + 3)^2 = 16$

Voila! We've successfully transformed our equation into the standard form. Notice how the equation now clearly shows the center and radius of the circle. The completing the square technique is a powerful tool, and mastering it will make your life much easier when dealing with circles and other conic sections.

Radius of the Circle Calculating the Distance

Now that we have the equation in standard form, $(x - 5)^2 + (y + 3)^2 = 16$, finding the radius is a piece of cake! Remember, the standard form is:

$(x - h)^2 + (y - k)^2 = r^2$

Where r is the radius. By comparing our equation with the standard form, we can see that:

r^2 = 16

To find r, we simply take the square root of both sides:

r = \sqrt{16}

r = 4

Therefore, the radius of the circle is 4 units. The radius of the circle is the distance from the center of the circle to any point on the circle's edge. In this case, we've determined that this distance is 4 units.

Key Insight: The standard form of the circle equation makes it incredibly easy to read off the radius. It's like having a treasure map where the radius is clearly marked!

Center of the Circle Identifying the Coordinates

While we were primarily focused on finding the radius, let's also take a moment to identify the center of the circle. Again, looking at the standard form equation:

$(x - 5)^2 + (y + 3)^2 = 16$

We can directly identify the center (h, k) by comparing it with the standard form $(x - h)^2 + (y - k)^2 = r^2$.

In our case:

• h = 5
• k = -3 (Note the sign change since the equation has (y + 3), which is equivalent to (y - (-3)))

So, the center of the circle is (5, -3). The center of the circle is the midpoint of any diameter of the circle. Knowing the center is crucial for various geometric calculations and constructions.

Quick Recap: From the standard form, we can easily extract both the radius and the center of the circle. It's like having a complete picture of the circle just by looking at its equation!

Visualizing the Circle Graphing the Equation

To truly grasp what we've calculated, let's visualize the circle. We know the center is (5, -3) and the radius is 4. Imagine plotting the center on a graph and then drawing a circle that extends 4 units in all directions from that center. You'll get a perfect circle that represents the equation we've been working with.

Graphing the equation helps solidify our understanding. It connects the abstract algebraic representation (the equation) with a concrete geometric shape (the circle). This visual connection is invaluable for problem-solving and deeper comprehension.

Practical Applications Real-World Examples

You might be wondering,