Let the vertices of a triangle are `A=(-3+2sin theta, 4+2 cos theta),` and `B=(-3+2cos theta, 4-2 cos theta)`, then the distance between the centroid and the circumcentre of `DeltaABC` is

Options

A. `(2)/(3)` units
B. `(3)/(2)` units
C. `(1)/(2)` units
D. `(1)/(3)` units

Answer

To find the distance between the centroid and the circumcenter of triangle ABC with vertices A and B given in the problem, we can follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are given as:
- \( A = (-3 + 2 \sin \theta, 4 + 2 \cos \theta) \)
- \( B = (-3 + 2 \cos \theta, 4 - 2 \cos \theta) \)
- We need to assume a third vertex \( C \) for the triangle. However, since the circumcenter can be determined from A and B, we can proceed with just these two points.

### Step 2: Find the circumcenter
The circumcenter of a triangle formed by points A and B can be found using the midpoint of the line segment connecting A and B. The circumcenter is the center of the circumcircle that passes through all three vertices.

1. **Midpoint of AB**:
\[ M = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) \]
Substituting the coordinates of A and B:
\[ M = \left( \frac{(-3 + 2 \sin \theta) + (-3 + 2 \cos \theta)}{2}, \frac{(4 + 2 \cos \theta) + (4 - 2 \cos \theta)}{2} \right) \]
Simplifying:
\[ M = \left( -3 + \frac{2 \sin \theta + 2 \cos \theta}{2}, 4 \right) = \left( -3 + \sin \theta + \cos \theta, 4 \right) \]

2. **Circumcenter (O)**:
Since the circumcenter lies on the perpendicular bisector of the segment AB, we can find it using the average of the coordinates of A and B. However, from the problem, we can directly see that the circumcenter is at:
\[ O = (-3, 4) \]

### Step 3: Find the centroid of the triangle
The centroid \( G \) of a triangle with vertices \( A, B, C \) is given by:
\[ G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right) \]
Assuming \( C \) is at the same position as \( A \) or \( B \) (since we are not provided with C), we can simplify by using A and B:
\[ G = \left( \frac{(-3 + 2 \sin \theta) + (-3 + 2 \cos \theta) + (-3)}{3}, \frac{(4 + 2 \cos \theta) + (4 - 2 \cos \theta) + 4}{3} \right) \]
This simplifies to:
\[ G = \left( \frac{-9 + 2 \sin \theta + 2 \cos \theta}{3}, \frac{12}{3} \right) = \left( -3 + \frac{2}{3}(\sin \theta + \cos \theta), 4 \right) \]

### Step 4: Calculate the distance between the centroid and circumcenter
The distance \( d \) between the circumcenter \( O(-3, 4) \) and the centroid \( G \) can be calculated using the distance formula:
\[ d = \sqrt{(x_G - x_O)^2 + (y_G - y_O)^2} \]
Substituting the coordinates:
\[ d = \sqrt{\left(-3 + \frac{2}{3}(\sin \theta + \cos \theta) + 3\right)^2 + (4 - 4)^2} \]
This simplifies to:
\[ d = \sqrt{\left(\frac{2}{3}(\sin \theta + \cos \theta)\right)^2} = \frac{2}{3} \sqrt{(\sin \theta + \cos \theta)^2} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ d = \frac{2}{3} \]

### Final Result
Thus, the distance between the centroid and the circumcenter of triangle ABC is:
\[ \boxed{\frac{2}{3}} \]