Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the circle satisfying the given conditions. Center \((4,3)\), goes through \((6,2)\)

Step-by-Step Solution

Answer

The equation of the circle is \((x - 4)^2 + (y - 3)^2 = 5\).

Step 1: Understand the Circle Equation

The standard equation of a circle is given by \[(x - h)^2 + (y - k)^2 = r^2\]where \((h, k)\) is the center of the circle and \(r\) is the radius.

Step 2: Substitute the Center into the Equation

Given that the center of the circle is \((4, 3)\), substitute \(h = 4\) and \(k = 3\) into the circle's equation. This gives us:\[(x - 4)^2 + (y - 3)^2 = r^2\]

Step 3: Find the Radius

We know that the circle passes through the point \((6, 2)\). Substitute \(x = 6\) and \(y = 2\) into the equation to find \(r\):\[(6 - 4)^2 + (2 - 3)^2 = r^2 \2^2 + (-1)^2 = r^2 \4 + 1 = r^2 \r^2 = 5\]So, the radius \(r\) is \(\sqrt{5}\).

Step 4: Write the Equation of the Circle

Now that we have the center and the radius, substitute \(r^2 = 5\) back into the equation from Step 2:\[(x - 4)^2 + (y - 3)^2 = 5\]

Key Concepts

Radius of a Circle

The radius of a circle is the distance from its center to any point on its circumference. It is an essential part of the circle equation. Understanding how to calculate the radius when given certain points is crucial.

To find the radius, you need two things:

• The coordinates of the center of the circle, which can be any point denoted as \((h, k)\).
• A point through which the circle passes, called the circumference point, denoted as \((x, y)\).

The distance formula can be used to calculate the radius. It is given by:\[r = \sqrt{(x - h)^2 + (y - k)^2}\]In our exercise, we know the center of the circle is \((4, 3)\) and it passes through \((6, 2)\). Applying the distance formula, the radius \(r\) is:\[r = \sqrt{(6 - 4)^2 + (2 - 3)^2} = \sqrt{4 + 1} = \sqrt{5}\]Thus, the radius of the circle is \(\sqrt{5}\). This tells us how wide the circle is from center to edge.

Center of a Circle

The center of a circle is a key component that defines the position of a circle in the coordinate plane. It is denoted by the coordinates \((h, k)\) and is directly used in the circle's equation.

The role of the center is to provide a consistent point from which measurements such as the radius can be calculated. In our specific case, the center is located at \((4, 3)\). This means:

• \(h = 4\) implies the circle is shifted 4 units along the x-axis.
• \(k = 3\) implies the circle is shifted 3 units upwards along the y-axis.

When substituting the center into a circle's equation, it helps simplify the calculation of other circle properties. Knowing the center ensures we have a fixed reference point to use in other related geometric calculations, such as determining the circumference or the diameter, which is twice the radius.

Standard Form of Circle Equation

The standard form of a circle's equation is a fundamental expression that includes both its center and its radius.

The equation is written as:\[(x - h)^2 + (y - k)^2 = r^2\]In this equation:

• \((h, k)\) represents the center of the circle.
• \(r\) is the radius of the circle.

This equation allows us to easily identify the circle’s major attributes: its center and radius. Our exercise gives the circle's center as \((4, 3)\) and the radius as \(\sqrt{5}\).

Substituting these values into the standard form equation, we find:\[(x - 4)^2 + (y - 3)^2 = 5\]This equation precisely describes our circle with its center and radius included, making it a powerful tool in both algebraic and geometric contexts. Understanding this equation is crucial for analyzing and graphing circles on a coordinate plane.