Problem 26 - 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. \((0,9)\) and \((0,-9)\)

Step-by-Step Solution

Answer

The center-radius form is \(x^2 + y^2 = 81\).

Step 1: Find the Midpoint

To find the center of the circle, calculate the midpoint of the diameter using the formula for the midpoint:\[\text{Midpoint} = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\]Given the endpoints \((0,9)\) and \((0,-9)\), substitute the values into the formula:\[\left(\frac{0+0}{2}, \frac{9+(-9)}{2}\right) = (0,0)\]Thus, the center of the circle is \((0,0)\).

Step 2: Calculate the Radius

Next, find the radius of the circle, which is the distance from the center to one of the endpoints of the diameter, \((0,9)\) or \((0,-9)\). Since both points are equidistant from the center, calculate using the point \((0,9)\). The formula for distance is:\[\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]Substituting the coordinates for the center \((0,0)\) and an endpoint \((0,9)\), we have:\[\sqrt{(0-0)^2 + (9-0)^2} = \sqrt{81} = 9\]The radius of the circle is 9.

Step 3: Write the Equation in Center-Radius Form

Finally, use the center-radius form of the equation of a circle:\[(x-h)^2 + (y-k)^2 = r^2\]where \((h,k)\) is the center and \(r\) is the radius. We found the center \((h,k)\) as \((0,0)\) and the radius \(r = 9\). Substitute these into the equation:\[(x-0)^2 + (y-0)^2 = 9^2\]which simplifies to:\[x^2 + y^2 = 81\]

Key Concepts

Midpoint Formula

The Midpoint Formula is a fundamental concept in geometry, particularly useful when dealing with lines and circles. It allows us to find the point that is exactly halfway between two given points. This midpoint is crucial in circle problems because it helps establish the center when given the endpoints of a diameter.

The formula is: - \[ \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \] - Where \(x_1, y_1\) and \(x_2, y_2\) are the coordinates of the endpoints.

By using this formula on the endpoints of a circle's diameter, like in our example with points \((0,9)\) and \((0,-9)\), we can determine the circle's center at \((0,0)\). This simplifies our subsequent calculations and forms a critical step in forming a circle's equation.

Center-Radius Form

The Center-Radius Form is an equation that succinctly represents a circle in a coordinate plane. It emphasizes the circle's center and its radius, making it straightforward to understand and utilize.
- The general formula is: \[ (x-h)^2 + (y-k)^2 = r^2 \] - - Here, \((h, k)\) is the center of the circle. - \(r\) represents the radius.

In practice, substituting the correct values for \((h, k)\) and \(r\) directly gives us the circle's equation. From our example, having found the center at \((0,0)\) and radius as \(9\), the equation becomes \(x^2 + y^2 = 81\).

This concise representation helps easily determine the circle's characteristics, facilitating further geometrical analysis or graphical depiction.

Distance Formula

The Distance Formula is another essential tool in geometry, used to calculate the distance between two points on the Cartesian plane. It is based on the Pythagorean Theorem and is particularly helpful when finding the radius of a circle given its center and a point on its perimeter.

The formula is: - \[\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\] - - Here, \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.
- From the original problem, by applying this formula to the center at \((0,0)\) and one endpoint of the diameter, \((0,9)\), we calculated a radius of \(9\). - This inequality provides a rigorous and quantifiable method to determine how large a circle is, which is often necessary for solving problems or deducing properties related to the circle.