Problem 43 - College Algebra Essentials

Question

give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation's domain and range. $$ (x-3)^{2}+(y-1)^{2}=36 $$

Step-by-Step Solution

Answer

The center of the circle is at (3,1), and the radius is 6 units. The domain of the circle is from -3 to 9, and the range is from -5 to 7.

Step 1: Identify the center

In the equation \((x-3)^{2}+(y-1)^{2}=36\), the values inside the parentheses with x and y represent the center of the circle. So the center is at (3,1).

Step 2: Find the radius

The right-hand side of the equation is the square of the radius, \(r^{2}\). In this case, \(r^{2} = 36\), so the radius can be found by taking the square root of 36, which gives \(r = 6\). So the radius of the circle is 6 units.

Step 3: Plot the center and radius on the graph

On a graph, mark the point (3,1), which is the center. From there, count 6 units in each direction to sketch in the circle, since this is the radius.

Step 4: Identify the domain and range

The domain is all possible x-values of the graph. The leftmost point of the circle is at \(x = 3 - 6 = -3\), and the rightmost point is at \(x = 3 + 6 = 9\). Therefore, the domain is from -3 to 9. Similarly, the range is all possible y-values of the graph. The lowest point of the circle is at \(y = 1 - 6 = -5\), and the highest point is at \(y = 1 + 6 = 7\). So, the range of the function is from -5 to 7.

Key Concepts

Center of a Circle

To understand equations of circles, knowing how to identify the center is crucial. The standard form equation of a circle in the coordinate plane is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) represents the center of the circle. In other words, the center is the point from which all points on the circle are equidistant. This makes circle geometry straightforward once you know the center.

For the equation \((x-3)^2 + (y-1)^2 = 36\), the terms \((x-3)\) and \((y-1)\) tell us that the center of the circle is at \((3, 1)\).

• The x-coordinate of the center is derived from \(x-3\), where 3 is the h value.
• The y-coordinate comes from \(y-1\), where 1 is the k value.

So, you simply take the numbers that x and y are subtracted from, and that gives you the center coordinates.

Radius of a Circle

In the equation of a circle, the radius is the distance from the center to any point on the circumference. It's signified by \(r\), and is found after equating the circle's equation form to that of \((x-h)^2 + (y-k)^2 = r^2\). Here, \(r^2\) represents the radius squared.

Examining the given equation, \((x-3)^2 + (y-1)^2 = 36\), 36 is actually \(r^2\). Hence, to find the radius, you need the square root of 36. That makes \(r = 6\).

• The radius dictates the circle's size and expands equally in all directions from the center.
• A larger radius results in a bigger circle and vice versa.

Knowing the radius is key in graphing and understanding the circle's spatial boundaries.

Domain and Range

Circle equations also help us determine the domain and range of the circle, which denote the limits of x and y values, respectively, covered by the circle.

The domain refers to the set of all x-values that the circle can cover on a graph. For our circle centered at \((3, 1)\) with a radius of 6, the furthest left the circle reaches is at an x-value of \(3 - 6 = -3\), and the furthest right is at \(3 + 6 = 9\). Therefore, its domain is \([-3, 9]\).

• Domain concerns the horizontal extent of the circle.

The range covers the span of y-values. Our circle's lowest point sits at \(y = 1 - 6 = -5\), and the highest at \(y = 1 + 6 = 7\). Thus, the range is \([-5, 7]\).

• Range outlines the circle's vertical scope.

Understanding the domain and range is vital for knowing how a shape occupies space on a graph.