Problem 106 - Intermediate Algebra

Question

Write the equation of a circle with a diameter whose endpoints are at \((-5,4)\) and \((7,-3)\)

Step-by-Step Solution

Answer

The equation of the circle is \((x-1)^2 + (y-0.5)^2 = \frac{193}{4}\).

Step 1: Understand the Circle Equation

The general equation of a circle with center at \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). To write this equation, we need to find the center \((h, k)\) and the radius \(r\).

Step 2: Find the Center of the Circle

Since the diameter's endpoints are given at \((-5,4)\) and \((7,-3)\), the center of the circle is the midpoint of the diameter. Use the midpoint formula: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\). Calculate: \(\left( \frac{-5 + 7}{2}, \frac{4 + (-3)}{2} \right) = \left( 1, \frac{1}{2} \right)\). So, the center is at \((1, 0.5)\).

Step 3: Calculate the Radius

The radius is half the length of the diameter. Use the distance formula to find the length of the diameter: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Substitute the points: \(\sqrt{(7 - (-5))^2 + (-3 - 4)^2} = \sqrt{12^2 + (-7)^2} = \sqrt{144 + 49} = \sqrt{193}\). The radius \(r = \frac{\sqrt{193}}{2}\).

Step 4: Substitute Values into the Circle Equation

Now that we have the center \((1, 0.5)\) and radius \(r = \frac{\sqrt{193}}{2}\), substitute these into the circle equation: \((x - 1)^2 + (y - 0.5)^2 = \left(\frac{\sqrt{193}}{2}\right)^2\). Simplify to obtain the equation of the circle: \((x-1)^2 + (y - 0.5)^2 = \frac{193}{4}\).

Key Concepts

Midpoint Formula

The midpoint formula helps you find the center point between two given points on a plane. It's particularly useful in geometry problems where you need to find the center of a line segment, such as a circle's diameter. The formula is:\[\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]To use the midpoint formula, simply add the x-coordinates of the two endpoints of the diameter together and divide by two. Do the same for the y-coordinates. This gives you the average position along both the x and y axes, which corresponds to the center of the line segment.

• Example: For the points \((-5,4)\) and \((7,-3)\), the midpoint is calculated as follows:
• Find the average of the x-coordinates: \( \frac{-5 + 7}{2} = 1 \)
• Find the average of the y-coordinates: \( \frac{4 + (-3)}{2} = 0.5 \)
• Thus, the midpoint and the center of the circle is \((1, 0.5)\).

Understanding the midpoint formula makes it easier to solve problems involving line segments and circle equations.

Distance Formula

The distance formula is a mathematical tool used to determine the length between two points in a two-dimensional plane. It is derived from the Pythagorean theorem and is especially handy for finding the length of a line segment, such as a circle's diameter.The formula is:\[\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]To apply this formula, subtract the x-coordinates of your two points, square the result, and do likewise for the y-coordinates. Then, add these two values and find the square root of the sum.

• Example: For endpoints \((-5,4)\) and \((7,-3)\):
• Calculate the difference in x-coordinates: \(7 - (-5) = 12\)
• Calculate the difference in y-coordinates: \((-3) - 4 = -7\)
• Use the formula: \(\sqrt{12^2 + (-7)^2} = \sqrt{193}\)
• Thus, the length of the diameter is \(\sqrt{193}\).

Mastering the distance formula can help with many geometric and coordinate plane problems.

Radius of a Circle

The radius of a circle is the distance from the center of the circle to any point on its circumference. When the diameter is known, finding the radius is straightforward, as the radius is simply half the length of the diameter.Formula:\[\text{Radius} = \frac{\text{Diameter}}{2}\]Applying this concept means you take the length you calculated for the diameter using the distance formula and divide by two to find the radius.

• Example: For a diameter length of \(\sqrt{193}\):
• Calculate the radius: \(\frac{\sqrt{193}}{2}\)
• Thus, the radius of the circle is \(\frac{\sqrt{193}}{2}\).

Knowing how to find the radius is key for writing the circle's equation or solving for other circle-related properties.