Finding The General Equation Of A Circle Passing Through Given Points

October 16, 2025 by Scholario Team 70 views

Hey guys! Today, we're diving deep into the fascinating world of circles and their equations. Specifically, we're going to tackle the challenge of finding the general equation of a circle when given certain conditions. This might sound intimidating, but trust me, we'll break it down step-by-step so you can master it. We'll explore two main scenarios: first, when the circle passes through two points and its center lies on the x-axis; and second, when the circle's center lies on the bisector of odd quadrants and passes through a given point.

1. Finding the General Equation of Circle λ Passing Through Points A(3, 1) and B(6, 2) with Center C on the x-axis

Let's start with the first scenario. Imagine we have a circle, which we'll call λ, that gracefully passes through two specific points, A(3, 1) and B(6, 2). But there's a catch! We also know that the center of this circle, which we'll call C, lies somewhere on the x-axis. Our mission, should we choose to accept it, is to find the general equation of this circle. How do we do it? Well, let's put on our mathematical thinking caps and get started!

Understanding the Basics of a Circle's Equation

Before we jump into the nitty-gritty, let's quickly recap the general equation of a circle. Remember, the general equation of a circle is expressed as:

x² + y² + 2gx + 2fy + c = 0

Where:

(-g, -f) represents the coordinates of the center of the circle.
The radius, r, can be found using the formula: r = √ (g² + f² - c)

This equation is our roadmap. If we can figure out the values of g, f, and c, we've essentially cracked the code and found the equation of our circle.

Leveraging the Given Information

Now, let's use the information we have. We know that the circle passes through points A(3, 1) and B(6, 2). This means that these points must satisfy the general equation of the circle. In simpler terms, if we plug in the x and y coordinates of these points into the equation, the equation should hold true. This gives us two equations:

For point A(3, 1):

3² + 1² + 2g(3) + 2f(1) + c = 0

Which simplifies to:

9 + 1 + 6g + 2f + c = 0

Or:

6g + 2f + c = -10 (Equation 1)
For point B(6, 2):

6² + 2² + 2g(6) + 2f(2) + c = 0

Which simplifies to:

36 + 4 + 12g + 4f + c = 0

Or:

12g + 4f + c = -40 (Equation 2)

We're making progress! We now have two equations with three unknowns (g, f, and c). But don't worry, we have another crucial piece of information: the center C lies on the x-axis.

The Center on the x-axis

What does it mean for the center C(-g, -f) to lie on the x-axis? It means that the y-coordinate of the center must be zero. In other words:

-f = 0

Therefore:

f = 0

This is a major breakthrough! We've just found the value of f. Now we can simplify our equations.

Solving for g and c

Substituting f = 0 into Equation 1 and Equation 2, we get:

From Equation 1:

6g + c = -10 (Equation 3)
From Equation 2:

12g + c = -40 (Equation 4)

Now we have a system of two equations with two unknowns (g and c). We can solve this using various methods, such as substitution or elimination. Let's use elimination. Subtracting Equation 3 from Equation 4, we get:

(12g + c) - (6g + c) = -40 - (-10)

6g = -30

g = -5

Great! We've found the value of g. Now we can substitute g = -5 back into either Equation 3 or Equation 4 to find c. Let's use Equation 3:

6(-5) + c = -10

-30 + c = -10

c = 20

We did it! We've found the values of g, f, and c.

The General Equation

Now that we have g = -5, f = 0, and c = 20, we can plug these values back into the general equation of the circle:

x² + y² + 2(-5)x + 2(0)y + 20 = 0

Simplifying, we get the general equation of the circle λ:

x² + y² - 10x + 20 = 0

And that's it! We've successfully found the general equation of the circle that passes through points A(3, 1) and B(6, 2) and has its center on the x-axis. Awesome work, guys!

2. Finding the Center of the Circle λ on the Bisector of Odd Quadrants Passing Through Points A(2, 8)

Now, let's tackle the second scenario. This time, we have a circle λ whose center C lies on the bisector of odd quadrants. This bisector is essentially the line y = x. We also know that this circle passes through a point A(2, 8). Our goal is to find the center of this circle. Ready for the challenge? Let's go! This problem steps up the complexity a bit, but with our problem-solving skills sharpened, we'll approach it with confidence.

Deciphering the Bisector of Odd Quadrants

The key to unlocking this problem lies in understanding what it means for the center C to lie on the bisector of odd quadrants. The bisector of odd quadrants is the line where the x and y coordinates are equal. Mathematically, this is represented by the equation y = x. This is a critical piece of information because it tells us that the coordinates of the center C must be of the form (a, a), where 'a' is some value.

Let's denote the center of the circle as C(a, a). This simplifies our problem significantly because now we only have one unknown variable to find.

The Distance is Key: Radius of the Circle

The next important concept to remember is that the distance from the center of the circle to any point on the circle is constant and equal to the radius (r). We know that the circle passes through point A(2, 8). Therefore, the distance between C(a, a) and A(2, 8) is the radius of the circle.

We can use the distance formula to express this mathematically:

r = √[(x₂ - x₁)² + (y₂ - y₁)²]

In our case, (x₁, y₁) = (a, a) and (x₂, y₂) = (2, 8). So, the radius is:

r = √[(2 - a)² + (8 - a)²]

This equation represents the radius of the circle in terms of 'a'. We're getting closer to our solution!

The Circle Equation: Another Perspective

Another way to think about the radius is to use the standard form of the equation of a circle, which is:

(x - h)² + (y - k)² = r²

Where (h, k) is the center of the circle and r is the radius. In our case, (h, k) = (a, a), so the equation becomes:

(x - a)² + (y - a)² = r²

Since the point A(2, 8) lies on the circle, it must satisfy this equation. Substituting x = 2 and y = 8, we get:

(2 - a)² + (8 - a)² = r²

Notice that this equation also involves r² and 'a'. This is a crucial connection that will allow us to solve for 'a'.

Solving for 'a': Putting It All Together

Now we have two expressions for r²:

From the distance formula: r² = (2 - a)² + (8 - a)²
From the circle equation: r² = (2 - a)² + (8 - a)²

Wait a minute... these are actually the same equation! This means we don't have two independent equations, which might seem like a problem. However, the equation itself holds the key to solving for 'a'.

Let's expand the equation:

(2 - a)² + (8 - a)² = r²

(4 - 4a + a²) + (64 - 16a + a²) = r²

2a² - 20a + 68 = r²

Now, remember that we are trying to find the center of the circle, (a, a). While we have an expression involving the radius squared, we realize that focusing on directly solving for 'a' is the most efficient approach. The fact that the point A(2,8) lies on the circle with center (a,a) gives us a crucial relationship to exploit.

Let's think back to the distance formula. The distance between the center (a, a) and the point (2, 8) is the radius. We can express the square of this distance (which is r²) as:

r² = (2 - a)² + (8 - a)²

Expanding this, as we did before, gives us:

r² = 2a² - 20a + 68

This is where the clever part comes in. To find the center (a, a), we need a specific value for 'a'. Let's think about what a circle is. It's the set of all points equidistant from the center. The equation we derived captures this relationship. However, to pinpoint 'a', we need to leverage another aspect of the problem, or perhaps re-interpret the information we already have.

Often in geometry problems, there's a hidden constraint or a slightly different way to view the given information. In this case, let's go back to the fact that the center lies on the line y = x and the circle passes through (2, 8). This means the distance between (a, a) and (2, 8) must be the same in all directions relative to the circle. We've already used the distance formula, but let's consider what this implies about the possible values of 'a'.

If we think about the geometry, there might be two possible circles that satisfy these conditions (a smaller circle and a larger circle). This suggests that the quadratic equation we derived might have two solutions for 'a'.

To find these solutions, we need to manipulate the equation into a more standard quadratic form, or perhaps consider a different approach altogether. Let's revisit the expansion of the squared terms and see if we can spot a way to simplify the equation to isolate 'a'.

2a² - 20a + 68 = r²

This equation relates 'a' and r². To solve for 'a', we ideally need to eliminate r² or find another independent equation. Since we only have one point on the circle and the center's location constraint, directly eliminating r² isn't straightforward.

However, let's consider a geometric insight. The perpendicular bisector of the line segment joining any two points on a circle passes through the center of the circle. We have one point on the circle, (2, 8), and we know the center lies on y = x. While we don't have a second point explicitly, we can use the concept of the perpendicular bisector.

The slope of the line segment joining (2, 8) and (a, a) is (8 - a) / (2 - a). The slope of the perpendicular bisector would be the negative reciprocal, which is (a - 2) / (8 - a). The midpoint of the line segment is ((2 + a)/2, (8 + a)/2). The perpendicular bisector must pass through this midpoint AND have the slope (a - 2) / (8 - a).

The equation of the perpendicular bisector would be:

y - (8 + a)/2 = [(a - 2) / (8 - a)] [x - (2 + a)/2]

Since the center (a, a) lies on this line, we can substitute x = a and y = a:

a - (8 + a)/2 = [(a - 2) / (8 - a)] [a - (2 + a)/2]

This looks complex, but let's simplify it:

(2a - 8 - a) / 2 = [(a - 2) / (8 - a)] [(2a - 2 - a) / 2]

(a - 8) / 2 = [(a - 2) / (8 - a)] [(a - 2) / 2]

Now we can cancel the /2 on both sides:

(a - 8) = [(a - 2)² / (8 - a)]

Multiply both sides by (8 - a):

(a - 8)(8 - a) = (a - 2)²

-(a - 8)² = (a - 2)²

-(a² - 16a + 64) = a² - 4a + 4

-a² + 16a - 64 = a² - 4a + 4

Now, let's move everything to one side to get a standard quadratic equation:

2a² - 20a + 68 = 0

Divide by 2:

a² - 10a + 34 = 0

Now we have a clean quadratic equation to solve for 'a'. We can use the quadratic formula:

a = [-b ± √(b² - 4ac)] / 2a

Where a = 1, b = -10, and c = 34

a = [10 ± √((-10)² - 4 * 1 * 34)] / 2

a = [10 ± √(100 - 136)] / 2

a = [10 ± √(-36)] / 2

Here, we encounter a negative value under the square root, which means the solutions for 'a' are complex numbers. This suggests that there might be no real circle that satisfies the given conditions. The geometry implies that the point (2, 8) is too far