Finding The Equation Of A Circle Given Two Points
Hey everyone! Let's dive into the fascinating world of circles and tackle a common problem in coordinate geometry: finding the equation of a circle when given certain conditions. In this article, we're going to break down a specific problem where we need to find the equation of a circle that touches another circle at a given point and also passes through another point. This type of problem might seem tricky at first, but with a systematic approach and a clear understanding of the underlying concepts, you'll be solving these in no time!
Understanding the Basics of Circles
Before we jump into the problem, let's quickly recap the essential concepts about circles. The standard equation of a circle is given by (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r is the radius. Knowing the center and radius is key to defining a circle's equation. When faced with a problem, our main goal is often to find these two crucial pieces of information. Additionally, remember that a point lies on the circle if and only if its coordinates satisfy the circle's equation. This simple fact is a powerful tool in solving many circle-related problems. You'll often find yourself substituting coordinates into the equation to form equations that help you solve for the unknowns h, k, and r. Another key concept is the distance formula, derived from the Pythagorean theorem, which helps us find the distance between two points. This is particularly useful for finding the radius of a circle when you know the center and a point on the circle. The tangent to a circle is a line that touches the circle at only one point, and the radius drawn to this point is always perpendicular to the tangent. This property is very important when dealing with circles that touch each other or touch lines. Understanding these fundamental properties is crucial for tackling more complex problems involving circles.
Problem Statement: A Circle Touching Another
Now, let's consider the specific problem we're going to solve. We are tasked with finding the equation of a circle that touches the circle given by the equation x^2 + y^2 - 6x + 2y + 5 = 0 at the point (4, -3) and also passes through the point (0, 7). This problem combines several concepts, making it a great exercise in applying our knowledge of circles. First, we have one circle touching another, which gives us a condition related to the centers and radii of the circles. The fact that they touch at a single point means that the line joining their centers passes through this point of tangency. This is a crucial geometric insight. Second, we have another point through which the circle must pass. This gives us another condition that we can use to form an equation. By combining these two pieces of information, we can set up a system of equations that will allow us to solve for the unknowns. The challenge lies in translating the geometric conditions into algebraic equations and then solving them. It's like piecing together a puzzle, where each condition provides a piece that helps us reveal the final picture, which in this case is the equation of the circle. This type of problem not only tests our understanding of circles but also our ability to think strategically and connect different mathematical concepts.
Step-by-Step Solution: Finding the Circle's Equation
Let's break down the solution step-by-step to make it easier to follow. First, we need to determine the center and radius of the given circle, x^2 + y^2 - 6x + 2y + 5 = 0. To do this, we'll rewrite the equation in the standard form (x - h)^2 + (y - k)^2 = r^2 by completing the square. Guys, remember how completing the square works? We group the x terms and the y terms, and then add and subtract the square of half the coefficient of x and y, respectively. So, we rewrite the equation as (x^2 - 6x) + (y^2 + 2y) + 5 = 0. Completing the square, we get (x^2 - 6x + 9) + (y^2 + 2y + 1) + 5 - 9 - 1 = 0, which simplifies to (x - 3)^2 + (y + 1)^2 = 5. Now we can clearly see that the center of the given circle is (3, -1) and its radius is √5.
Next, let's denote the center of the circle we want to find as (h, k) and its radius as r. Since the two circles touch at (4, -3), the distance between their centers is equal to the sum or difference of their radii. This is because the centers and the point of tangency are collinear. The distance between (3, -1) and (h, k) is √((h - 3)^2 + (k + 1)^2). So, we have √((h - 3)^2 + (k + 1)^2) = r ± √5. Additionally, since the point (4, -3) lies on the circle we're trying to find, it must satisfy the equation (x - h)^2 + (y - k)^2 = r^2. Substituting the coordinates of the point, we get (4 - h)^2 + (-3 - k)^2 = r^2. The fact that the point (0, 7) also lies on the circle gives us another equation: (0 - h)^2 + (7 - k)^2 = r^2.
Now we have a system of three equations with three unknowns (h, k, and r). Solving this system might seem daunting, but don't worry, we'll take it step by step. We have:
- √((h - 3)^2 + (k + 1)^2) = r ± √5
- (4 - h)^2 + (-3 - k)^2 = r^2
- h^2 + (7 - k)^2 = r^2
From equations 2 and 3, we can equate the expressions for r^2, giving us (4 - h)^2 + (-3 - k)^2 = h^2 + (7 - k)^2. Expanding and simplifying this equation will eliminate r and give us a relationship between h and k. This is a crucial step as it reduces the complexity of the system. Once we have a relationship between h and k, we can substitute it back into one of the original equations to solve for one variable in terms of the other. Then, we can use the first equation to solve for r. This might involve some algebraic manipulation and careful simplification, but with patience and attention to detail, we can find the values of h, k, and r. Once we have these values, we can finally write the equation of the circle in the standard form. Remember, the key is to break down the problem into smaller, manageable steps and use the given information strategically to form equations that lead us to the solution.
Algebraic Manipulation and Solving for h, k, and r
Let's continue with solving the system of equations. From equations 2 and 3, we have:
(4 - h)^2 + (-3 - k)^2 = h^2 + (7 - k)^2
Expanding both sides, we get:
16 - 8h + h^2 + 9 + 6k + k^2 = h^2 + 49 - 14k + k^2
Notice that the h^2 and k^2 terms cancel out, which simplifies the equation significantly. Now, let's rearrange the terms to get a linear equation in h and k:
-8h + 6k + 25 = 49 - 14k
-8h + 20k = 24
Dividing by 4, we further simplify the equation to:
-2h + 5k = 6
This gives us a relationship between h and k. We can express h in terms of k (or vice versa). Let's express h in terms of k:
2h = 5k - 6
h = (5k - 6) / 2
Now we have an expression for h in terms of k. We can substitute this into equation 3 (or equation 2, it doesn't matter) to get an equation involving only k and r. Substituting into equation 3, we have:
((5k - 6) / 2)^2 + (7 - k)^2 = r^2
Expanding and simplifying this equation will give us another relationship between k and r. This step involves a bit more algebraic manipulation, but it's crucial to eliminate one more variable. We get:
(25k^2 - 60k + 36) / 4 + (49 - 14k + k^2) = r^2
Multiplying through by 4 to eliminate the fraction:
25k^2 - 60k + 36 + 196 - 56k + 4k^2 = 4r^2
29k^2 - 116k + 232 = 4r^2
Now we have an equation relating k and r^2. We still need to use equation 1, which involves the square root, to bring in the condition about the circles touching each other. Remember equation 1:
√((h - 3)^2 + (k + 1)^2) = r ± √5
Substitute h = (5k - 6) / 2 into this equation:
√(((5k - 6) / 2 - 3)^2 + (k + 1)^2) = r ± √5
Simplifying the expression inside the square root:
√(((5k - 12) / 2)^2 + (k + 1)^2) = r ± √5
√((25k^2 - 120k + 144) / 4 + k^2 + 2k + 1) = r ± √5
Multiplying through by 4 inside the square root:
√(25k^2 - 120k + 144 + 4k^2 + 8k + 4) / 2 = r ± √5
√(29k^2 - 112k + 148) / 2 = r ± √5
Now, square both sides to eliminate the square root:
(29k^2 - 112k + 148) / 4 = (r ± √5)^2
29k^2 - 112k + 148 = 4(r^2 ± 2r√5 + 5)
We have two possible cases here, one with r + √5 and one with r - √5. Let's consider the case with r + √5 first. We'll substitute the expression for 4r^2 from the previous equation:
29k^2 - 112k + 148 = 29k^2 - 116k + 232 ± 8r√5 + 20
We can see that the 29k^2 terms cancel out, which is a good sign. This simplifies the equation further. Now we have a linear equation in k and r√5. Solving this equation along with the previous equation relating k and r^2 will give us the values of k and r. Once we have k, we can find h using the expression h = (5k - 6) / 2.
This process involves careful algebraic manipulation and attention to detail, but by breaking it down into smaller steps, we can systematically solve for the unknowns. Remember, the key is to use the given information strategically and to form equations that help us eliminate variables and simplify the problem. It might seem like a lot of work, but the satisfaction of finding the solution is well worth the effort!
Final Steps and the Circle's Equation
After simplifying and substituting, we eventually arrive at the values for h, k, and r. This part often involves solving a quadratic equation or a system of linear equations. Once we have these values, we can plug them back into the standard equation of a circle, (x - h)^2 + (y - k)^2 = r^2, to get the final equation of the circle. It's a great feeling when you finally see the equation that represents the circle we've been working so hard to find!
It's also a good idea to double-check your answer. You can do this by plugging the given points, (4, -3) and (0, 7), into the equation to make sure they satisfy it. This helps ensure that you haven't made any mistakes along the way. Additionally, you can verify that the distance condition between the centers and the radii is satisfied. This comprehensive check gives you confidence that your solution is correct. And there you have it! We've successfully navigated through the problem, using our knowledge of circles and some clever algebraic techniques to find the equation of the circle. These types of problems might seem challenging at first, but with practice and a clear understanding of the concepts, you'll become a pro at solving them.
Conclusion: Mastering Circle Equations
Finding the equation of a circle given specific conditions is a fundamental skill in coordinate geometry. Problems like this one, where we have a circle touching another circle and passing through a given point, require a good understanding of circle properties and algebraic manipulation. By breaking down the problem into smaller steps, we can systematically solve for the unknowns and arrive at the final equation. Remember, the key is to translate the geometric conditions into algebraic equations and then use techniques like completing the square and solving systems of equations to find the center and radius of the circle. Practice is essential for mastering these skills. The more you practice, the more comfortable you'll become with the concepts and the techniques involved. So, keep practicing, keep exploring, and keep enjoying the beauty of mathematics!
This article walked through a detailed example, showing how to approach these problems step-by-step. Remember, understanding the underlying principles is key. Keep practicing, and you'll become a circle-solving expert in no time! Guys, keep up the great work, and happy solving!