The Radius and Equation of a Circle Given Endpoints of a Diameter

In this article, we explore how to find the radius and equation of a circle given the endpoints of its diameter. We'll walk through the steps using the given endpoints -4, 5 and 2, -3. This example will help you understand the process for similar problems, whether you're a student or a professional needing to work with geometric shapes.

1. Understanding the Mathematics Behind the Problem

The problem provides two endpoints of a diameter, -4, 5 and 2, -3. A key property of a circle is that its diameter is a line segment with the endpoints lying on the circle and passing through the center. The length of the diameter is crucial for determining the circle's properties, including its radius and equation.

1.1 Calculating the Center of the Circle

The center of the circle can be determined using the mid-point of the diameter endpoints. The mid-point formula is given by: [text{Mid-point} left( frac{x_1 x_2}{2}, frac{y_1 y_2}{2} right)] Substituting the given points -4, 5 and 2, -3 into the formula, we get:

[text{Mid-point} left( frac{-4 2}{2}, frac{5 - 3}{2} right) (-1, 1)]

Thus, the center of the circle is at (-1, 1).

1.2 Calculating the Radius of the Circle

The radius is half the length of the diameter. To find the length of the diameter, we use the distance formula:

[text{Distance} sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2}]

Substituting the coordinates (-4, 5) and (2, -3) into the formula, we get:

[text{Diameter} sqrt{(2 - (-4))^2 (-3 - 5)^2} sqrt{(2 4)^2 (-8)^2} sqrt{6^2 8^2} sqrt{36 64} sqrt{100} 10]

Since the diameter is 10, the radius ( r ) is half of the diameter:

[text{Radius} frac{10}{2} 5]

1.3 Deriving the Equation of the Circle

The general form of the equation of a circle is given by: [(x - h)^2 (y - k)^2 r^2] where ((h, k)) is the center of the circle and (r) is the radius. Substituting (h -1), (k 1), and (r 5) into this equation, we get:

[(x 1)^2 (y - 1)^2 25]

2. Graphical Solution

For a graphical solution, we can visualize the circle using the center and radius. The midpoint of the two endpoints of the diameter gives us the center at (-1, 1). The radius is 5 units. Visualizing the circle, we see that it passes through both endpoints (-4, 5) and (2, -3).

3. Verification

To verify the solution, we can check if the given endpoints satisfy the circle's equation.

Using endpoint (-4, 5) and substituting into the circle's equation:

[text{Left Side} (-4 1)^2 (5 - 1)^2 (-3)^2 4^2 9 16 25]

Since the left side equals the right side (25 25), the point (-4, 5) lies on the circle.

Similarly, using endpoint (2, -3):

[text{Left Side} (2 1)^2 (-3 - 1)^2 3^2 (-4)^2 9 16 25]

Again, the left side equals the right side (25 25), verifying that the point (2, -3) also lies on the circle.

4. Key Takeaways

1. The center of the circle can be found using the midpoint formula of the endpoints of the diameter. 2. The radius is half the length of the diameter, which can be calculated using the distance formula. 3. The general form of the circle's equation is ((x - h)^2 (y - k)^2 r^2), where ((h, k)) is the center and (r) is the radius.

Understanding these steps and formulas will help you solve similar problems efficiently. If you encounter more complex problems, the principles remain the same.

Keywords: circle equation, radius calculation, circle endpoints