Problem 22 - 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. $$(-4,5) \text { and }(6,-9)$$

Step-by-Step Solution

Answer

The circle's equation is \((x - 1)^2 + (y + 2)^2 = 74\).

Step 1: Find the Midpoint of the Diameter

The midpoint of the diameter is the center of the circle. Use the midpoint formula: \((x_1, y_1)\) and \((x_2, y_2)\) are the endpoints of the diameter. The formula is: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\). For the points \((-4, 5)\) and \((6, -9)\), the midpoint is \(\left( \frac{-4 + 6}{2}, \frac{5 + (-9)}{2} \right) = (1, -2)\). This is the center of the circle.

Step 2: Calculate the Radius

The radius is the distance from the center of the circle to any one of the endpoints of the diameter. Use the distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Using center \((1, -2)\) and one endpoint \((-4, 5)\), the radius is \(\sqrt{(1 - (-4))^2 + (-2 - 5)^2} = \sqrt{(5)^2 + (-7)^2} = \sqrt{25 + 49} = \sqrt{74}\).

Step 3: Write the Equation in Center-Radius Form

The center-radius form of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Using center \((1, -2)\) and radius \(\sqrt{74}\), the equation is \((x - 1)^2 + (y + 2)^2 = 74\).

Key Concepts

Midpoint Formula

The midpoint formula is a handy tool for finding the center point between two coordinates. When you have two endpoints,

• \((x_1, y_1)\)
• \((x_2, y_2)\)

the midpoint is calculated using the formula: \[\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]This formula averages the x-coordinates and y-coordinates to find the point exactly halfway between.

In the exercise, the endpoints \((-4, 5)\) and \((6, -9)\) give us a midpoint \((1, -2)\). This midpoint isn't just any point; it's the center of the circle as the circle's diameter runs through these endpoints.

Center-Radius Form

The center-radius form of a circle equation is integral when defining the circle. It uses specific parameters to make the equation clear and straightforward.

The form is given by:\[(x - h)^2 + (y - k)^2 = r^2\] where:

• \((h, k)\) is the center of the circle
• \(r\) is the radius of the circle

This form makes it easy to quickly see both the center and radius of the circle directly from the equation.

Using the values from our exercise, the center is \((1, -2)\) and the radius is \(\sqrt{74}\), giving us the circle's equation as \((x-1)^2 + (y+2)^2 = 74\). This representation helps in visualizing and graphing circles accurately.

Distance Formula

The distance formula is essential for finding the length between two points in the coordinate plane. This is particularly useful to determine the radius of a circle when you know the circle's center and a point on its circumference.

The formula is expressed as:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This calculates the straight-line distance, or the length of the hypotenuse in a right triangle formed by the two points.

In the exercise, the center \((1, -2)\) and the endpoint \((-4, 5)\) are used to find the radius. Plugging into the formula yields:\[\sqrt{(1 - (-4))^2 + (-2 - 5)^2} = \sqrt{(5)^2 + (-7)^2} = \sqrt{25 + 49} = \sqrt{74}\]This shows the elegant way the formula helps in determining the exact distance, which in this case is the radius of the circle.