Problem 46 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 35-46: Find an equation of the circle that satisfies the stated conditions. $$ \text { Endpoints of a diameter } A(-5,2) \text { and } B(3,6) $$

Step-by-Step Solution

Answer

The equation of the circle is \((x + 1)^2 + (y - 4)^2 = 20\).

Step 1: Find the Center of the Circle

The center of a circle that has a diameter as one of its line segments is the midpoint of that diameter. First, identify the coordinates of the endpoints of the diameter, which are given as \( A(-5,2) \) and \( B(3,6) \). Then calculate the midpoint using the midpoint formula: \( \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \). Substitute the known values: \( \text{Midpoint} = \left( \frac{-5 + 3}{2}, \frac{2 + 6}{2} \right) = \left( -1, 4 \right) \). Therefore, the center of the circle is \( (-1, 4) \).

Step 2: Calculate the Radius of the Circle

The radius of the circle is half the length of the diameter. First, find the distance between the endpoints \( A(-5,2) \) and \( B(3,6) \) using the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). This gives \( d = \sqrt{(3 + 5)^2 + (6 - 2)^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} \). Therefore, the radius is \( \frac{\sqrt{80}}{2} = \sqrt{20} = 2\sqrt{5} \).

Step 3: Write the Equation of the Circle

Now that you have the center \((-1, 4)\) and the radius \(2\sqrt{5}\), use the standard equation of a circle, \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Substitute the known values: \((x + 1)^2 + (y - 4)^2 = (2\sqrt{5})^2\). Simplify \((2\sqrt{5})^2\) to get 20, thus the equation becomes \((x + 1)^2 + (y - 4)^2 = 20\).

Key Concepts

Midpoint Formula

To find the center of a circle when you have the endpoints of the diameter, you use the midpoint formula. This is because the midpoint of the diameter of a circle is the circle's center. The formula is:

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

Substituting the coordinates of the endpoints can help find the midpoint accurately. Here, with endpoints \( A(-5,2) \) and \( B(3,6) \), the midpoint is calculated as follows:

• Midpoint: \( \left( \frac{-5 + 3}{2}, \frac{2 + 6}{2} \right) = (-1, 4) \).

Thus, the circle's center is at \( (-1, 4) \). This simple formula makes it easy to find the circle's center. It's just averaging the x-coordinates and the y-coordinates separately.

Distance Formula

The distance formula is crucial for calculating the length of the diameter or any line segment in a coordinate plane. It essentially extends the Pythagorean theorem to two points in the plane. The formula is:

• Distance: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Using endpoints \( A(-5,2) \) and \( B(3,6) \), the distance between these points is calculated as:

• \( d = \sqrt{(3 + 5)^2 + (6 - 2)^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} \).

The distance is part of determining the diameter's length. By dividing the result by 2, we find the radius, \( \frac{\sqrt{80}}{2} = 2\sqrt{5} \). This radius is essential for the circle's equation.

Standard Equation of a Circle

The standard equation of a circle is written using the circle's center and radius. It is expressed as:

• Equation: \( (x - h)^2 + (y - k)^2 = r^2 \)

Here, \((h, k)\) represent the circle's center, and \(r\) is the radius. Based on our earlier calculations:

• Center: \((-1, 4)\)
• Radius: \(2\sqrt{5}\)

The equation becomes:

• \((x + 1)^2 + (y - 4)^2 = (2\sqrt{5})^2\)
• Simplified: \((x + 1)^2 + (y - 4)^2 = 20\)

This formula encodes all necessary information about the circle's position and size.

Geometry

Geometry is the mathematical study of shapes, sizes, and properties of space. A circle is a significant shape in geometry, defined as the set of all points equidistant from a fixed point, known as the center. Understanding circles involves several key concepts:

• The center is a crucial reference point, found using the midpoint of a diameter.
• The radius is the distance from the center to any point on the circle, often found using the distance formula.
• The diameter, being twice the radius, provides a full measure across the circle through the center.

The equations and formulas used in circle geometry help calculate precise attributes of circles, aiding in solving complex geometric problems effectively. Grasping these foundational principles is essential for further exploration in geometry and related fields.