Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 equation of the circle is \((x-2)^2 + (y-5)^2 = 5\).

Step 1: Find the midpoint of the diameter

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

Step 2: Calculate the length of the diameter

Use the distance formula to find the length of the diameter \(AB\), which is given by: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Substituting the points \(A = (1,3)\) and \(B = (3,7)\), the length is \(d = \sqrt{(3 - 1)^2 + (7 - 3)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\).

Step 3: Find the radius of the circle

Since the diameter \(AB\) is \(2\sqrt{5}\), the radius \(r\) is half of the diameter. Therefore, \(r = \frac{2\sqrt{5}}{2} = \sqrt{5}\).

Step 4: Write the equation of the circle

The general equation of a circle with center \((h,k)\) and radius \(r\) is given by \((x-h)^2 + (y-k)^2 = r^2\). Here, the center \((h,k) = (2,5)\) and \(r = \sqrt{5}\). Therefore, the equation of the circle is \((x-2)^2 + (y-5)^2 = 5\).

Key Concepts

Midpoint Formula

To find the center of a circle when given the endpoints of its diameter, the midpoint formula is essential. The formula helps determine the midpoint, which acts as the center of the circle. The formula is given by:

• If the endpoints are \(A = (x_1, y_1)\) and \(B = (x_2, y_2)\), then the midpoint \(M \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\).

For example, with points \(A = (1,3)\) and \(B = (3,7)\), calculate the midpoint to find:

• \(M = \left( \frac{1+3}{2}, \frac{3+7}{2} \right) = (2, 5)\).

This midpoint, \(M = (2, 5)\), serves as the circle's center. Understanding how to calculate the midpoint is key to locating the exact position of the circle's center from its diameter.

Distance Formula

The distance formula allows you to compute the distance between two points in a plane. It is especially useful to determine the length of a circle's diameter when you have the endpoints. The formula is expressed as:

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

For the endpoints \(A = (1,3)\) and \(B = (3,7)\), plug these into the formula:

• \(d = \sqrt{(3 - 1)^2 + (7 - 3)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\).

This gives a distance of \(2\sqrt{5}\), which is the diameter of the circle. This calculated distance is vital for further determining the radius.

Radius of a Circle

The radius of a circle is a crucial component of its definition and is used to write the circle's equation. It can be thought of as half the length of the diameter. Once you know the diameter, finding the radius is straightforward:

• For a diameter length \(d\), the radius \(r = \frac{d}{2}\).

In our specific case, where the diameter \(d = 2\sqrt{5}\), the radius \(r = \frac{2\sqrt{5}}{2} = \sqrt{5}\).

• The radius informs not just the size, but its application in the circle's equation as \(r^2\).

Understanding the determination of the radius from the diameter forms the foundation for describing a circle algebraically.

Diameter and Radius Relationship

The diameter and radius of a circle are fundamentally linked. The diameter is twice the length of the radius. Or conversely, the radius is half of the diameter.

• This relationship is expressed as: diameter \(d = 2r\) and radius \(r = \frac{d}{2}\).

So, knowing one directly helps find the other. For example, if a circle's diameter is \(2\sqrt{5}\), then its radius \(r = \sqrt{5}\).

• This simple yet powerful relationship is key when you set up equations involving circles in coordinate geometry.

Comprehending this connection helps solve problems involving the circle's equation, ensuring that each component of the circle is correctly integrated into finding and mapping its characteristics.