The centre of the circle `x=2+3 cos theta, y=3 sin theta-1`, is

Options

A. (3, 3)
B. (2, -1)
C. (-2, 1)
D. (-1, 2)

Explanation

To find the center of the circle given by the parametric equations \( x = 2 + 3 \cos \theta \) and \( y = 3 \sin \theta - 1 \), we can follow these steps:

### Step 1: Rewrite the Parametric Equations
We start with the given equations:
\[ x = 2 + 3 \cos \theta \]
\[ y = 3 \sin \theta - 1 \]

### Step 2: Solve for \(\cos \theta\) and \(\sin \theta\)
From the equation for \(x\), we can isolate \(\cos \theta\):
\[ \cos \theta = \frac{x - 2}{3} \]
From the equation for \(y\), we can isolate \(\sin \theta\):
\[ \sin \theta = \frac{y + 1}{3} \]

### Step 3: Use the Pythagorean Identity
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting the expressions we found for \(\sin \theta\) and \(\cos \theta\):
\[ \left(\frac{y + 1}{3}\right)^2 + \left(\frac{x - 2}{3}\right)^2 = 1 \]

### Step 4: Simplify the Equation
Multiplying through by \(9\) (which is \(3^2\)) to eliminate the denominators:
\[ (y + 1)^2 + (x - 2)^2 = 9 \]

### Step 5: Identify the Center and Radius
The equation \((x - 2)^2 + (y + 1)^2 = 9\) is in the standard form of a circle:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius. Here, we can identify:
- \(h = 2\)
- \(k = -1\)
- \(r^2 = 9\) (so \(r = 3\))

### Step 6: Write the Center of the Circle
Thus, the center of the circle is:
\[ (2, -1) \]

### Final Answer
The center of the circle is \((2, -1)\).

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