To find the parametric equations for the given equation [tex]\((x+1)^2 + (y-3)^2 = 16\)[/tex], let's follow a detailed, step-by-step approach:
1. Identify the type of curve represented by the equation:
The given equation [tex]\((x+1)^2 + (y-3)^2 = 16\)[/tex] is in the standard form of the equation of a circle:
[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.
2. Determine key parameters:
By comparing [tex]\((x + 1)^2 + (y - 3)^2 = 16\)[/tex] to the standard form:
[tex]\[
h = -1, \quad k = 3, \quad r^2 = 16
\][/tex]
We find that:
[tex]\[
h = -1, \quad k = 3, \quad r = \sqrt{16} = 4
\][/tex]
3. Recall the parametric equations for a circle:
For a circle centered at [tex]\((h, k)\)[/tex] with radius [tex]\(r\)[/tex], the parametric equations are:
[tex]\[
x = h + r \cos(t)
\][/tex]
[tex]\[
y = k + r \sin(t)
\][/tex]
4. Substitute the values for [tex]\(h\)[/tex], [tex]\(k\)[/tex], and [tex]\(r\)[/tex]:
Substituting [tex]\(h = -1\)[/tex], [tex]\(k = 3\)[/tex], and [tex]\(r = 4\)[/tex]:
[tex]\[
x = -1 + 4 \cos(t) \quad \text{or} \quad x = 4 \cos(t) - 1
\][/tex]
[tex]\[
y = 3 + 4 \sin(t) \quad \text{or} \quad y = 4 \sin(t) + 3
\][/tex]
5. Combine the parametric equations:
Thus, the parametric equations for the given circle are:
[tex]\[
x = 4 \cos(t) - 1
\][/tex]
[tex]\[
y = 4 \sin(t) + 3
\][/tex]
6. Check the provided options:
Comparing this with the provided options, we see:
[tex]\[
\text{Option 3: } x = 4 \cos(t) - 1 \text{ and } y = 4 \sin(t) + 3
\][/tex]
Hence, the correct parametric equations that represent [tex]\((x+1)^2 + (y-3)^2 = 16\)[/tex] are:
[tex]\[
x = 4 \cos(t) - 1 \quad \text{and} \quad y = 4 \sin(t) + 3
\][/tex]
[tex]\(\boxed{x = 4 \cos (t) - 1 \text { and } y = 4 \sin (t) + 3}\)[/tex]
