Unveiling Circle Radii: A Geometric Quest

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Unveiling Circle Radii: A Geometric Quest

Hey math enthusiasts! Ever found yourselves tangled in a web of circles, points, and equations? Today, we're diving headfirst into a classic geometry problem: finding the radius of a circle that gracefully dances through a given point and the dramatic intersection points of two other circles. Sounds fun, right? Buckle up, because we're about to embark on a mathematical adventure that'll test our skills and hopefully leave you feeling like geometry rockstars.

The Grand Setup: Our Geometric Playground

Alright, let's set the stage. We're given a specific point, (3, 1), which is like our guiding star. Then, we have two circles that are doing their own thing: (x - 2)^2 + (y - 1)^2 = 5 and (x - 1)^2 + (y - 2)^2 = 1. Our mission, should we choose to accept it (and we definitely should!), is to discover the radius of a new circle. This new circle is a real social butterfly; it has to waltz through our given point (3, 1) and hit up the exact spots where our two existing circles decide to mingle. Think of it as finding the perfect circle that connects these specific locations. Before we proceed to the solution, let's establish a clear understanding of the fundamental concepts that will be employed in our approach.

First, let's explore the fundamental properties of circles. The equation of a circle is typically expressed in the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the coordinates of the circle's center, and r denotes its radius. This equation essentially encapsulates all the points that lie at a fixed distance (the radius) from the center. Understanding this is crucial because it gives us a direct link between the geometric properties of a circle and its algebraic representation. Secondly, understanding the intersection of circles is very important. The points of intersection of two circles are the solutions to the system of equations formed by their individual equations. These intersection points are the places where the circles meet, and they satisfy the equations of both circles. In order to actually find the points of intersection, we can solve the system of equations. There are several ways to do this: substitution, elimination, or, if we're feeling particularly ambitious, using geometric reasoning. In our case, we'll probably lean on algebra to do the heavy lifting.

Now, let's tackle the meat of the problem: finding the radius. Imagine we have these two circles, doing their thing, and they cross each other in two points. Our mystery circle has to pass exactly through those two points, and also through (3, 1). To get the radius, we'll need to know the center of our new circle. Think about it: once we know the center and any point on the circle, finding the radius is just a simple distance calculation. So, the game plan is: first, find the points where the two circles intersect. Then, use those intersection points along with (3, 1) to figure out the center of our new circle. Finally, we can use the distance formula to calculate the radius itself. We're essentially going on a treasure hunt, using equations as our maps, and the radius is the gold we seek. So, let's roll up our sleeves and get started!

Unveiling the Intersection Points: Where Circles Meet

Alright, let's get our hands dirty and find those elusive intersection points. We have the equations of our two circles: (x - 2)^2 + (y - 1)^2 = 5 and (x - 1)^2 + (y - 2)^2 = 1. The challenge is to find the values of x and y that satisfy both equations simultaneously. This is the classic system of equations scenario.

One approach is to use the method of substitution or elimination. Here's a breakdown of how we could use elimination:

  1. Expand the equations: Expand both equations to get them into a more manageable form.
    • Equation 1: x^2 - 4x + 4 + y^2 - 2y + 1 = 5, which simplifies to x^2 - 4x + y^2 - 2y = 0.
    • Equation 2: x^2 - 2x + 1 + y^2 - 4y + 4 = 1, which simplifies to x^2 - 2x + y^2 - 4y = -4.
  2. Eliminate a variable: Notice that both equations have x^2 and y^2 terms. Subtracting the second equation from the first will help eliminate these terms. So, (x^2 - 4x + y^2 - 2y) - (x^2 - 2x + y^2 - 4y) = 0 - (-4), which simplifies to -2x + 2y = 4.
  3. Solve for one variable: From -2x + 2y = 4, we can easily solve for y: y = x + 2.
  4. Substitute back: Substitute y = x + 2 into either of the original equations. Let's use the simplified version of Equation 1: x^2 - 4x + (x + 2)^2 - 2(x + 2) = 0.
  5. Simplify and solve the quadratic: Expanding and simplifying gives us x^2 - 4x + x^2 + 4x + 4 - 2x - 4 = 0, which simplifies further to 2x^2 - 2x = 0. Factoring out 2x gives us 2x(x - 1) = 0. Therefore, x = 0 or x = 1.
  6. Find the corresponding y values: If x = 0, then y = 0 + 2 = 2. If x = 1, then y = 1 + 2 = 3.

So, the intersection points are (0, 2) and (1, 3). We now know exactly where our target circle needs to pass through!

Crafting the New Circle's Equation: The Equation's Tale

Now we're in the home stretch, guys! We've found the intersection points (0, 2) and (1, 3), and we have our guiding star, the point (3, 1). Armed with this information, we're going to build the equation of our new circle. Remember, the general form is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. We just need to figure out h, k, and r.

Here’s how we can tackle this:

  1. General Equation: Since our circle passes through the intersection points (0, 2) and (1, 3), and the given point (3, 1), we can set up the following equations:

    • For (0, 2): (0 - h)^2 + (2 - k)^2 = r^2 => h^2 + (2 - k)^2 = r^2
    • For (1, 3): (1 - h)^2 + (3 - k)^2 = r^2 => (1 - h)^2 + (3 - k)^2 = r^2
    • For (3, 1): (3 - h)^2 + (1 - k)^2 = r^2 => (3 - h)^2 + (1 - k)^2 = r^2

  2. Eliminating r^2: Notice that all three equations equal r^2. We can use this to eliminate r^2 and create a system of equations with just h and k.

    • Equation 1 & 2: h^2 + (2 - k)^2 = (1 - h)^2 + (3 - k)^2
    • Equation 2 & 3: (1 - h)^2 + (3 - k)^2 = (3 - h)^2 + (1 - k)^2

  3. Simplifying the Equations: Expand and simplify the equations above:

    • Equation 1 & 2: h^2 + 4 - 4k + k^2 = 1 - 2h + h^2 + 9 - 6k + k^2 => 2h + 2k = 6 => h + k = 3
    • Equation 2 & 3: 1 - 2h + h^2 + 9 - 6k + k^2 = 9 - 6h + h^2 + 1 - 2k + k^2 => 4h - 4k = 0 => h = k

  4. Solving for h and k: We now have a simple system of equations. Substitute h = k into h + k = 3. Therefore, k + k = 3 => 2k = 3 => k = 1.5. Since h = k, then h = 1.5.

  5. Finding r: Now that we know the center is (1.5, 1.5), we can plug these values, along with any of the three points, into the circle equation to find r^2. Let's use (3, 1):

    • (3 - 1.5)^2 + (1 - 1.5)^2 = r^2
    • (1.5)^2 + (-0.5)^2 = r^2
    • 2.25 + 0.25 = r^2
    • r^2 = 2.5
    • r = √2.5

Thus, the radius of the circle is √2.5, or approximately 1.58. We’ve done it, guys! We've successfully navigated the geometric maze and found our treasure: the radius of the circle that fits all the criteria!

The Grand Finale: Calculating the Radius

Alright, friends, we've done all the heavy lifting! We’ve found the center of our new circle to be (1.5, 1.5). Now, to find the radius, we just need to calculate the distance between the center and any of the points on the circle. Remember, the circle passes through (0, 2), (1, 3), and (3, 1). Let's use the point (3, 1) and the distance formula. The distance formula is essentially the Pythagorean theorem applied to coordinate geometry: distance = √((x2 - x1)^2 + (y2 - y1)^2).

  1. Apply the Distance Formula: Using the center (1.5, 1.5) and the point (3, 1), we get:

    r = √((3 - 1.5)^2 + (1 - 1.5)^2)

  2. Calculate the Differences: Subtract the coordinates:

    r = √((1.5)^2 + (-0.5)^2)

  3. Square the Differences: Square each of the differences:

    r = √(2.25 + 0.25)

  4. Add the Squares: Add the squared values:

    r = √2.5

  5. Find the Square Root: Take the square root to get the radius:

    r ≈ 1.58

So, the radius of the circle is approximately 1.58 units. We've done it! We started with a set of intersecting circles and a single point, and through a series of algebraic and geometric steps, we've successfully unveiled the radius of the circle that connects them all. High five, everyone! This journey highlights the elegant interplay of different mathematical concepts and the power of methodical problem-solving. This problem shows how seemingly complex geometric relationships can be understood and solved using fundamental mathematical tools. From understanding the equation of a circle to finding the intersection points of two circles, and finally, using the distance formula, we've built a solid foundation for tackling similar problems in the future. Remember, the key is to break down the problem into smaller, manageable steps, and always keep an eye on the big picture. Now go forth and conquer more geometric challenges!