Problem 43
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
Verified Answer
The center of the circle is at the point (3,1), with a radius of 6. The domain is [-3, 9] and the range is [-5, 7].
1Step 1: Identify the Center
The center of the circle can be directly read off from the standard form of the equation, by identifying values of h and k from \((x - h)^2 + (y - k)^2 = r^2\). In the given equation, \((x-3)^{2}+(y-1)^{2}=36\), \(h = 3\) and \(k = 1\) . Hence, the center of the circle is at the point \((3,1)\).
2Step 2: Identify the Radius
The radius of the circle can be found from the right side of the standard form equation, by taking the square root of \(r^2\). From the given equation, \(r^2\) is 36, hence the radius \(r\) is \(\sqrt{36}\) which simplifies to 6.
3Step 3: Graph the Circle
Sketch the graph of the circle by starting at the center point \((3,1)\) and making a circle with radius 6. This is done by counting 6 units right, left, up, and down from the center and connecting those points to form a circle.
4Step 4: Identify the Domain and Range
The domain (possible x-values) of the circle is the interval of all x-values hit by the circle, which in this case goes from the leftmost point of the circle to the rightmost. Since the center is at \(x = 3\) and the radius is 6, this interval ranges from \(3-6\) to \(3+6\), or \([-3, 9]\). Similarly, the range (possible y-values) goes from the bottom of the circle to the top. The center is at \(y = 1\) and radius is 6 so the range is from \(1-6\) to \(1+6\), or \([-5, 7]\).
Key Concepts
Center and Radius of a CircleStandard Form of a Circle EquationDomain and Range of a CircleGraphing a Circle
Center and Radius of a Circle
When you have the equation of a circle in the standard form \((x - h)^2 + (y - k)^2 = r^2\), the center and radius are easily determined. Here,
- \(h\) and \(k\) represent the x and y coordinates of the circle's center, which means from the equation, \((x-3)^{2}+(y-1)^{2}=36\), the center is at \((3, 1)\).
- \(r^2\) specifies the radius squared, so to get the radius \(r\), you just take the square root of 36, resulting in a radius of 6.
Standard Form of a Circle Equation
The standard form of a circle equation is a special form that makes understanding circle characteristics straightforward. This form is \((x - h)^2 + (y - k)^2 = r^2\), where:
- \(h\) denotes the x-coordinate of the circle's center,
- \(k\) denotes the y-coordinate of the circle's center,
- and \(r\) is the radius of the circle.
Domain and Range of a Circle
The domain and range of a circle help us understand the set of x and y values that the circle covers. For the circle given by \((x-3)^{2}+(y-1)^{2}=36\):
- The **domain** consists of all x-values the circle covers. The center is at \(x = 3\) and the radius is 6. Thus, the domain extends from \((3-6) = -3\) to \((3+6) = 9\). In interval notation, this is \([-3, 9]\).
- The **range** represents all y-values covered by the circle. The center being at \(y = 1\) with a radius of 6 means the range goes from \((1-6) = -5\) to \((1+6) = 7\). This range is denoted by \([-5, 7]\).
Graphing a Circle
Graphing circles can be done systematically by understanding the equation of the circle. First, locate the center of the circle using the form \((x - h)^2 + (y - k)^2 = r^2\). For the equation \((x-3)^{2}+(y-1)^{2}=36\), the center is \((3,1)\). Here are the steps to graph it accurately:
- Place a point at the center \((3, 1)\).
- From this center, count 6 units in all directions (up, down, left, right) which represents the radius.
- Connect these points in a smooth round shape. That’s your circle!
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