Problem 60 - Algebra and Trigonometry

Question

\(55-62\) . Find an equation of the circle that satisfies the given conditions. Endpoints of a diameter are \(P(-1,3)\) and \(Q(7,-5)\)

Step-by-Step Solution

Answer

The equation of the circle is \((x - 3)^2 + (y + 1)^2 = 32\).

Step 1: Find the Midpoint of the Diameter

The center of the circle is the midpoint of the diameter. Use the midpoint formula, which is given by \( M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \). For points \( P(-1, 3) \) and \( Q(7, -5) \), the midpoint is: \( M = \left( \frac{-1 + 7}{2}, \frac{3 - 5}{2} \right) = (3, -1) \). Thus, the center of the circle is at \( (3, -1) \).

Step 2: Find the Distance (Radius)

The radius is half of the length of the diameter. First, find the length of the diameter using the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \). Substituting the coordinates of \( P \) and \( Q \): \( d = \sqrt{(7 + 1)^2 + (-5 - 3)^2} = \sqrt{8^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \). Therefore, the radius \( r = \frac{8\sqrt{2}}{2} = 4\sqrt{2} \).

Step 3: Write the Equation of the Circle

The general form for the equation of a circle is \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h,k)\) is the center and \(r\) is the radius. Substituting the center \((3, -1)\) and radius \(4\sqrt{2}\) into the formula: \[(x - 3)^2 + (y + 1)^2 = (4\sqrt{2})^2\]. Simplifying: \[(x - 3)^2 + (y + 1)^2 = 32\].

Key Concepts

Understanding the Midpoint Formula

The midpoint formula is a useful tool in geometry that helps us find the exact center point between two given points in a coordinate plane. Imagine you have a line segment connecting two points, let's call them \( P(x_1, y_1) \) and \( Q(x_2, y_2) \).
The midpoint \( M \) of this segment can be calculated using the formula:

• \( M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)

This formula simply takes the average of the \( x \)-coordinates and the average of the \( y \)-coordinates. This provides a new point \( M \) that is exactly halfway between \( P \) and \( Q \). For example, if \( P = (-1, 3) \) and \( Q = (7, -5) \),
their midpoint is \((3, -1)\). This point \((3, -1)\) then becomes the center of our circle, because it's the midpoint of the diameter.

Exploring the Distance Formula

The distance formula helps us find the length of a line segment between two points in a coordinate plane. This is essential when trying to calculate the diameter or radius of a circle based on endpoints of a diameter.
Given two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \), the distance \( d \) between them is calculated as follows:

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

This formula is derived from the Pythagorean theorem and calculates the hypotenuse of a right triangle formed between the points. Applying this to our example with points \( P = (-1, 3) \) and \( Q = (7, -5) \), yields:
\( d = \sqrt{(7 + 1)^2 + (-5 - 3)^2} = \sqrt{8^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \).
This result of \( 8\sqrt{2} \) represents the diameter length of the circle.

Understanding the Diameter of a Circle

In a circle, the diameter is a line that extends from one point on the circle, passing through the center, to another point on the circle. It is essentially the longest distance across the circle. Generally, the diameter is twice the radius of the circle.
When you have endpoints of a diameter, you can use the distance formula to find the length of this diameter, as illustrated earlier with distance \( 8\sqrt{2} \).
The diameter tells us how wide the circle is, and it can also be used to find other properties of the circle. For example:

• The radius can be found by dividing the diameter by two.

This makes the diameter a fundamental concept in understanding the size and positioning of any circle based on given points.

Delving into the Radius of a Circle

The radius is the distance from the center of a circle to any point on its circumference. It is half the length of the diameter, making it a vital measurement in equations and understanding circle properties.
From our example, the diameter \( 8\sqrt{2} \) gets us the radius by simply dividing the diameter by two:

• \( r = \frac{8\sqrt{2}}{2} = 4\sqrt{2} \)

The radius is crucial for writing the equation of a circle. In the standard equation form \((x - h)^2 + (y - k)^2 = r^2\), the radius \( r \) squared is used.
For our circle with center \((3, -1)\) and radius \(4\sqrt{2}\), the equation becomes \((x - 3)^2 + (y + 1)^2 = 32\). Knowing the radius allows us to explore additional attributes of the circle, such as area, which is calculated as π times the square of the radius.