Finding The Circle Equation Through Points A(1, -6), B(2, 1), C(5, 2)
Hey guys! Let's dive into a fun math problem today: finding the general equation of a circle that passes through three given points. Specifically, we're looking for the circle that goes through A(1, -6), B(2, 1), and C(5, 2). This might sound tricky, but we'll break it down step-by-step to make it super clear. So, grab your pencils and let's get started!
1. Understanding the General Equation of a Circle
Before we jump into the calculations, let’s quickly refresh the general equation of a circle. The general equation of a circle is given by:
x² + y² + Dx + Ey + F = 0
Where D, E, and F are constants that we need to find. Each set of values for these constants will define a unique circle. Our goal here is to find the specific values of D, E, and F that correspond to the circle passing through our points A, B, and C. Think of it like solving a puzzle where each point gives us a clue to unlock the circle's equation. We're essentially reverse-engineering the circle from points on its edge! The cool thing is that once we have this equation, we know everything about the circle – its center, its radius, and all the points that lie on it.
Now, you might be wondering, why this particular form of the equation? Well, it's derived from the more familiar standard form of a circle's equation, which is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. The general form is what we get when we expand the squares and rearrange the terms. It's less intuitive for directly reading off the center and radius, but it's incredibly useful for situations like this, where we have points and need to find the equation. By plugging in the coordinates of our points A, B, and C into this general form, we'll create a system of equations that we can solve for D, E, and F. It's like turning geometry into algebra, which is a pretty neat trick!
2. Plugging in the Points
Now comes the fun part – plugging in the coordinates of our points A(1, -6), B(2, 1), and C(5, 2) into the general equation. This is where the magic happens, as each point will give us a unique equation. Let's start with point A(1, -6). Substituting x = 1 and y = -6 into the general equation, we get:
(1)² + (-6)² + D(1) + E(-6) + F = 0
Simplifying this, we have:
1 + 36 + D - 6E + F = 0
Which further simplifies to:
D - 6E + F = -37 (Equation 1)
See how we've transformed a point into an equation? Now, let's do the same for point B(2, 1). Substituting x = 2 and y = 1, we get:
(2)² + (1)² + D(2) + E(1) + F = 0
Which simplifies to:
4 + 1 + 2D + E + F = 0
And further simplifies to:
2D + E + F = -5 (Equation 2)
We're on a roll! One more point to go. For point C(5, 2), substituting x = 5 and y = 2, we get:
(5)² + (2)² + D(5) + E(2) + F = 0
Simplifying this, we get:
25 + 4 + 5D + 2E + F = 0
And finally:
5D + 2E + F = -29 (Equation 3)
Alright, guys, we've done the heavy lifting of plugging in the points. Now we have a system of three linear equations with three unknowns (D, E, and F). This is a classic algebra problem, and there are several ways we can solve it. We can use substitution, elimination, or even matrices. The key is to systematically eliminate variables until we can solve for each one individually. It might seem like a lot of work, but each step gets us closer to our goal – the equation of the circle!
3. Solving the System of Equations
Okay, now we're at the heart of the problem: solving the system of equations. We have three equations:
Equation 1: D - 6E + F = -37 Equation 2: 2D + E + F = -5 Equation 3: 5D + 2E + F = -29
There are a couple of ways we can tackle this. Let's use the elimination method, which is often a clean and efficient approach. The idea is to eliminate one variable at a time by adding or subtracting multiples of the equations. Notice that each equation has a single F term, which makes F a good candidate for elimination.
First, let's eliminate F from Equations 1 and 2. We can do this by subtracting Equation 1 from Equation 2:
(2D + E + F) - (D - 6E + F) = -5 - (-37)
This simplifies to:
D + 7E = 32 (Equation 4)
Great! We've created a new equation with only D and E. Now, let's eliminate F again, this time from Equations 1 and 3. Subtracting Equation 1 from Equation 3:
(5D + 2E + F) - (D - 6E + F) = -29 - (-37)
Simplifies to:
4D + 8E = 8
We can further simplify this by dividing the entire equation by 4:
D + 2E = 2 (Equation 5)
Now we have two equations (Equation 4 and Equation 5) with two unknowns (D and E):
Equation 4: D + 7E = 32 Equation 5: D + 2E = 2
Let's eliminate D by subtracting Equation 5 from Equation 4:
(D + 7E) - (D + 2E) = 32 - 2
This simplifies to:
5E = 30
Dividing by 5, we get:
E = 6
Fantastic! We've found E. Now we can plug this value back into either Equation 4 or Equation 5 to solve for D. Let's use Equation 5:
D + 2(6) = 2
D + 12 = 2
D = -10
Awesome, we've found D as well! Now that we have D and E, we can plug them back into any of our original three equations to solve for F. Let's use Equation 1:
(-10) - 6(6) + F = -37
-10 - 36 + F = -37
-46 + F = -37
F = 9
Woohoo! We've solved the system of equations! We found that D = -10, E = 6, and F = 9. This was a marathon of algebra, but we made it through! Now we're just one step away from the grand finale: writing the general equation of the circle.
4. Writing the General Equation
Alright, guys, we've reached the final stretch! We've successfully found the values of D, E, and F by solving the system of equations. Remember, these constants are the key to unlocking the general equation of the circle. We found that:
- D = -10
- E = 6
- F = 9
Now, we simply plug these values back into the general equation of a circle:
x² + y² + Dx + Ey + F = 0
Substituting our values, we get:
x² + y² + (-10)x + (6)y + 9 = 0
Which simplifies to:
x² + y² - 10x + 6y + 9 = 0
And there you have it! This is the general equation of the circle that passes through the points A(1, -6), B(2, 1), and C(5, 2). 🎉
We did it! We started with three points, dove into the general equation of a circle, set up a system of equations, and solved for the unknowns. It was quite a journey, but hopefully, you found it clear and understandable. Remember, math problems like these might seem daunting at first, but breaking them down into smaller, manageable steps makes them much easier to tackle. Keep practicing, and you'll become a pro at solving these kinds of problems in no time!
Conclusion
In this article, we walked through the process of finding the general equation of a circle given three points. We covered the basics of the general equation, how to plug in the points to create a system of equations, how to solve that system using elimination, and finally, how to write the equation. This is a fundamental concept in geometry and algebra, and mastering it opens the door to solving many other interesting problems. So, keep exploring, keep learning, and most importantly, keep having fun with math!
