Problem 40 - Algebra 2

Question

Write an equation for the circle that satisfies each set of conditions. endpoints of a diameter at (5, -9) and (3, 11)

Step-by-Step Solution

Answer

The circle equation is \((x - 4)^2 + (y - 1)^2 = 101.\)

Step 1: Identify Circle Properties

To write the equation of a circle, we need the center and the radius. The center is the midpoint of the endpoints of the diameter, and the radius is half of the distance between the endpoints.

Step 2: Find the Midpoint

The midpoint of a diameter with endpoints (\(x_1, y_1\)) and (\(x_2, y_2\)) can be found using the formula: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right).\)For the given points (5, -9) and (3, 11), \(\text{Midpoint} = \left( \frac{5 + 3}{2}, \frac{-9 + 11}{2} \right) = (4, 1).\)

Step 3: Calculate the Radius

The distance between the endpoints of the diameter can be calculated using the distance formula: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.\)So,\[\sqrt{(3 - 5)^2 + (11 + 9)^2} = \sqrt{(-2)^2 + (20)^2} = \sqrt{4 + 400} = \sqrt{404}.\]The radius is half of the diameter, so:\(\text{Radius} = \frac{\sqrt{404}}{2} = \frac{\sqrt{404}}{2}.\)

Step 4: Write the Equation of the Circle

With center \((4, 1)\) and radius \(\frac{\sqrt{404}}{2}\), the equation of the circle is:\[(x - 4)^2 + (y - 1)^2 = \left(\frac{\sqrt{404}}{2}\right)^2.\] Simplify the right side: \[\left(\frac{\sqrt{404}}{2}\right)^2 = \frac{404}{4} = 101.\]Thus, the equation is:\((x - 4)^2 + (y - 1)^2 = 101.\)

Key Concepts

Midpoint Formula

The midpoint formula is a key tool in geometry used to find the center point between two distinct coordinates. Imagine you are bridging a gap between two points, such as the endpoints of a line. The formula is ideal for finding the exact spot in the middle. This is sometimes called the "average" of the points.
The formula to find the midpoint between two points \( (x_1, y_1) \, \text{and}\, (x_2, y_2) \) is given by:

• \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

For example, with endpoints at \( (5, -9) \, \text{and}\, (3, 11) \), the calculation becomes:

• \( \left( \frac{5 + 3}{2}, \frac{-9 + 11}{2} \right) \).
• This yields \( (4, 1) \), the center of the circle found by considering the average of the x-coordinates and y-coordinates separately.

Recognizing the midpoint is crucial when working with diameters, as it helps you pinpoint the circle's center.

Distance Formula

The distance formula helps you determine how far apart two points are on a coordinate plane. Visualize it as the method to measure the length of the straight line connecting them, just like a ruler.
The distance formula between two points \( (x_1, y_1) \, \text{and}\, (x_2, y_2) \) is:

• \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Applying this to the endpoints \( (5, -9) \, \text{and}\, (3, 11) \), we have:

• \( \sqrt{(3 - 5)^2 + (11 - (-9))^2} \).
• Calculating the squared differences, you find \( \sqrt{(-2)^2 + (20)^2} \).
• This simplifies to \( \sqrt{404} \), indicating the diameter's length.

This calculation is essential in finding the length of the diameter, which is twice the length of the radius.

Circle Radius

The circle's radius is simply half the diameter's length. Think of the radius as the arm of a compass that draws the circle, reaching from the center to any point on the circumference.
Once we have the distance across the diameter, we can find the radius by dividing this distance by 2. For our points, we computed the diameter as \( \sqrt{404} \). Hence, the radius becomes:

• \( \frac{\sqrt{404}}{2} \).

Understanding the radius is vital because it determines how "big" or "small" the circle is and directly influences the circle's equation.

Diameter Endpoints

Endpoints of the diameter are key reference points defining the size of a circle. They lie exactly opposite each other across the circle's center.
In the context of a circle, the diameter is double the radius. Knowing the endpoints (

• such as (5, -9) and (3, 11)

means you have a direct line stretching from one edge of the circle to the other. This connection provides all the dimensional clues you need:

• It helps you find the circle's midpoint, by averaging these endpoint coordinates.
• It allows you to compute the radius, by measuring the full stretch of the diameter and then halving it.

Grasping these endpoints is crucial for forming the complete picture of the circle's geometry.