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Question 9 (Essay Worth 10 points)

(07.03 MC)

Part A: What is the equation of a circle with center [tex]\((-2, 4)\)[/tex] and a diameter of 6 units? Show all necessary steps. (4 points)

Part B: Explain how to graph the circle by hand on the coordinate plane. (3 points)

Part C: What is the domain of the circle? Explain how to determine the domain. (3 points)

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Question 10 (Essay Worth 10 points)

(07.03 MC)



Answer :

Part A: Equation of the Circle

To find the equation of a circle, we use the standard form of a circle's equation, which is given by:

[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]

where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.

1. Identify the center of the circle: The center of the circle is given as [tex]\((-2, 4)\)[/tex]. Therefore, [tex]\(h = -2\)[/tex] and [tex]\(k = 4\)[/tex].

2. Determine the diameter of the circle: The diameter of the circle is given as 6 units.

3. Calculate the radius: The radius is half of the diameter.
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{6}{2} = 3 \text{ units} \][/tex]

4. Substitute the center [tex]\((h, k)\)[/tex] and the radius [tex]\(r\)[/tex] into the standard equation:
[tex]\[ (x - (-2))^2 + (y - 4)^2 = 3^2 \][/tex]

5. Simplify the equation:
[tex]\[ (x + 2)^2 + (y - 4)^2 = 9 \][/tex]

Thus, the equation of the circle is:
[tex]\[ (x + 2)^2 + (y - 4)^2 = 9 \][/tex]

Part B: Graphing the Circle by Hand

To graph the circle on a coordinate plane:

1. Plot the center of the circle at [tex]\((-2, 4)\)[/tex].
2. From the center, measure the radius (3 units) in all four cardinal directions (up, down, left, and right).
- To the right: [tex]\((-2 + 3, 4) = (1, 4)\)[/tex]
- To the left: [tex]\((-2 - 3, 4) = (-5, 4)\)[/tex]
- Upward: [tex]\((-2, 4 + 3) = (-2, 7)\)[/tex]
- Downward: [tex]\((-2, 4 - 3) = (-2, 1)\)[/tex]
3. Plot these four points: [tex]\((1, 4)\)[/tex], [tex]\((-5, 4)\)[/tex], [tex]\((-2, 7)\)[/tex], and [tex]\((-2, 1)\)[/tex].
4. Draw a smooth curve connecting these points to form the circle. Make sure the curve is equidistant from the center at all points.

Part C: Domain of the Circle

The domain of the circle refers to the set of all possible [tex]\(x\)[/tex]-values that lie within the circle.

1. Recognize that for a circle, the domain is determined by the horizontal extent of the circle, from the leftmost point to the rightmost point.

2. The center of the circle is at [tex]\(x = -2\)[/tex] with a radius of 3 units.

3. Thus, the leftmost point is:
[tex]\[ x = -2 - 3 = -5 \][/tex]

4. The rightmost point is:
[tex]\[ x = -2 + 3 = 1 \][/tex]

5. Therefore, the domain of the circle is all [tex]\(x\)[/tex]-values between [tex]\(-5\)[/tex] and [tex]\(1\)[/tex] inclusive. The domain can be written as:
[tex]\[ \text{Domain: } [-5, 1] \][/tex]

So the final domain of the circle is [tex]\([-5, 1]\)[/tex].