Problem 24 - A Graphical Approach to College Algebra

Question

We can find an equation of a circle if we know the coordinates of the endpoints of a diameter of the circle. First, find the midpoint of the diameter, which is the center of the circle. Then find the radius, which is the distance from the center to either endpoint of the diameter. Finally use the center-radius form to find the equation. Find the center-radius form for each circle having the given endpoints of a diameter. $$(-3,-2) \text { and }(1,-4)$$

Step-by-Step Solution

Answer

The equation of the circle is \((x + 1)^2 + (y + 3)^2 = 5\).

Step 1: Find the Midpoint

The midpoint of the diameter of a circle is the center of the circle. To find the midpoint between the points \((-3, -2)\) and \((1, -4)\), we use the midpoint formula: \((x_m, y_m) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\). Substitute the given points into the formula: \[x_m = \frac{-3 + 1}{2} = \frac{-2}{2} = -1\] \[y_m = \frac{-2 + (-4)}{2} = \frac{-6}{2} = -3\] Thus, the midpoint, or center of the circle, is \((-1, -3)\).

Step 2: Calculate the Radius

The radius of the circle is the distance from the center to either endpoint of the diameter. Use the distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Substitute one endpoint and the center: \(x_1 = -1, y_1 = -3, x_2 = -3, y_2 = -2\). \[\text{Radius} = \sqrt{(-3 + 1)^2 + (-2 + 3)^2} = \sqrt{(-2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}\]

Step 3: Write the Circle Equation

With the center \((-1, -3)\) and radius \(\sqrt{5}\), apply the center-radius form of a circle's equation: \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) the radius. Substitute values into the equation: \((x + 1)^2 + (y + 3)^2 = (\sqrt{5})^2\). Simplify the equation: \((x + 1)^2 + (y + 3)^2 = 5\).

Key Concepts

Midpoint Formula

The idea behind the Midpoint Formula is quite simple, yet extremely useful. Imagine you have two points on a plane. The midpoint is a point right in the middle of these two, creating a balance between them. Think of it as the perfect line divider between the two points.

To find this midpoint, you use the formula:

• For the x-coordinate, add the x-coordinates of both points and divide by 2: \[ x_m = \frac{x_1 + x_2}{2} \]
• For the y-coordinate, do the same with the y-coordinates: \[ y_m = \frac{y_1 + y_2}{2} \]

In our original problem, the midpoint between (-3, -2) and (1, -4) was needed to find the circle's center. By substituting these coordinates into the formula, you get the center (-1, -3). This midpoint is crucial because it indicates the exact center from where the circle eventually unfolds.

Distance Formula

Whether you are hopping between cities, measuring lines, or finding radii of circles, the Distance Formula is your faithful companion. This formula helps you find the "straight-line" distance between any two points on a plane.

Here's how you calculate it:

• First, subtract the x-coordinates of the two points: \[ x_2 - x_1 \]
• Second, subtract the y-coordinates of the two points: \[ y_2 - y_1 \]
• Square both results: \[(x_2 - x_1)^2 \text{ and } (y_2 - y_1)^2\]
• Add these squares together and take the square root of the sum: \[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

In our example, by applying it to one endpoint and the center (-1, -3), you find a distance, or radius, of \( \sqrt{5} \). This distance is vital as it tells you how far the circle extends from the center.

Center-Radius Form

The Center-Radius Form of the circle’s equation is an elegant way to express circles. It takes the mystery out of equations, clearly showing where the circle is located and how big it is.

This form is written as:

• \[ (x - h)^2 + (y - k)^2 = r^2 \]
• Here, \((h, k)\) is the center of the circle, and \(r\) is the radius.

In our scenario, using the center (-1, -3) and the radius \( \sqrt{5} \), the center-radius equation becomes:

• \[ (x + 1)^2 + (y + 3)^2 = 5 \]

This equation shows all necessary details: the displacement of the circle from the origin and how large a space it covers. This form beautifully ties everything together, revealing the comprehensive nature of your circle.