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].
