Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(11-16\), find the equation of the circle satisfying the given conditions. Diameter \(A B\), where \(A=(1,3)\) and \(B=(3,7)\)

Step-by-Step Solution

Answer

The circle's equation is \((x-2)^2 + (y-5)^2 = 5\).

Step 1: Find the midpoint

The center of the circle is the midpoint of the diameter AB. To find the midpoint, use the formula: \[ M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \] For points A(1,3) and B(3,7): \[ M = \left(\frac{1+3}{2}, \frac{3+7}{2}\right) = (2, 5) \] So, the center of the circle is (2, 5).

Step 2: Calculate the radius

The radius is half of the length of the diameter. First, find the length of AB using the distance formula: \[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \] For A(1,3) and B(3,7): \[ d = \sqrt{(3-1)^2 + (7-3)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \] The radius \( r \) is half of \( \sqrt{20} \), so: \[ r = \frac{\sqrt{20}}{2} = \sqrt{5} \]

Step 3: Write the equation of the circle

The equation of a circle with center \( (h, k) \) and radius \( r \) is: \[ (x-h)^2 + (y-k)^2 = r^2 \] Here, \( h = 2 \), \( k = 5 \), and \( r = \sqrt{5} \). Plug these into the equation: \[ (x-2)^2 + (y-5)^2 = (\sqrt{5})^2 \] Simplify: \[ (x-2)^2 + (y-5)^2 = 5 \] This is the equation of the circle.

Key Concepts

Midpoint Formula

When dealing with circles, especially those defined by a diameter, the midpoint formula is an essential tool. It helps us find the center point of the circle. The midpoint is simply the average of the x-coordinates and the y-coordinates of the endpoints of a line segment. This makes it quite intuitive and straightforward.
The formula for the midpoint, given two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \), is: \[ M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \] Here’s how it works:

• Add the x-coordinates of the two points.
• Divide the sum by 2 to get the x-coordinate of the midpoint.
• Repeat the same steps for the y-coordinates.

This way, the midpoint \( M \) gives us the center of the diameter. Using the points \( A(1,3) \) and \( B(3,7) \), the center becomes \( (2, 5) \). This point will be our circle's center.

Distance Formula

The distance formula helps us find the length between two points in a plane. Understanding this formula is crucial for finding both the diameter and radius of a circle when its endpoints are given. The formula is an application of the Pythagorean Theorem, which relates the sides of a right triangle.
To calculate the distance \( d \) between two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \), we apply: \[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \] Here's a step-by-step:

• Subtract the x-coordinate of point A from the x-coordinate of point B.
• Do the same for the y-coordinates.
• Square both results.
• Add the squares together.
• Take the square root of the sum to find the distance.

For example, with points \( A(1,3) \) and \( B(3,7) \), the distance, which is the diameter here, computes to \( \sqrt{20} = 2\sqrt{5} \). Knowing this helps us find the circle's radius.

Circle Radius

The radius of a circle is a fundamental concept that defines its size. It is the distance from the center of the circle to any point on its circumference. In problems where the diameter is given, the radius is simply half that length.
Once you have the diameter, finding the radius is straightforward:

• Calculate the length of the diameter using the distance formula.
• Divide the diameter by 2 to find the radius.

In our example, the diameter between points A and B is computed as \( 2\sqrt{5} \). Halving this, we find that the radius \( r \) of our circle is \( \sqrt{5} \).
This radius is then used to write the circle's equation based on its center \( (2, 5) \) and the general formula for the equation of a circle: \[ (x-h)^2 + (y-k)^2 = r^2 \] This gives us the final circle equation, \( (x-2)^2 + (y-5)^2 = 5 \), with radius \( \sqrt{5} \).