Show that the centroid of the triangle with vertices `(5 cos theta, 4 sin theta), (4 cos theta, 5 sin theta)` and (0, 0) lies on the circle `x^(2)+y^(2)=9`.

Answer

To show that the centroid of the triangle with vertices \((5 \cos \theta, 4 \sin \theta)\), \((4 \cos \theta, 5 \sin \theta)\), and \((0, 0)\) lies on the circle defined by the equation \(x^2 + y^2 = 9\), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are given as:
- \(A(5 \cos \theta, 4 \sin \theta)\)
- \(B(4 \cos \theta, 5 \sin \theta)\)
- \(C(0, 0)\)

### Step 2: Calculate the coordinates of the centroid
The formula for the centroid \(G\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]

Substituting the coordinates of the vertices:
- \(x_1 = 5 \cos \theta\)
- \(y_1 = 4 \sin \theta\)
- \(x_2 = 4 \cos \theta\)
- \(y_2 = 5 \sin \theta\)
- \(x_3 = 0\)
- \(y_3 = 0\)

The \(x\)-coordinate of the centroid is:
\[ G_x = \frac{5 \cos \theta + 4 \cos \theta + 0}{3} = \frac{9 \cos \theta}{3} = 3 \cos \theta \]

The \(y\)-coordinate of the centroid is:
\[ G_y = \frac{4 \sin \theta + 5 \sin \theta + 0}{3} = \frac{9 \sin \theta}{3} = 3 \sin \theta \]

Thus, the coordinates of the centroid \(G\) are:
\[ G(3 \cos \theta, 3 \sin \theta) \]

### Step 3: Verify if the centroid lies on the circle
To check if the centroid lies on the circle defined by \(x^2 + y^2 = 9\), we substitute \(G_x\) and \(G_y\) into the equation of the circle:
\[ x^2 + y^2 = (3 \cos \theta)^2 + (3 \sin \theta)^2 \]

Calculating this gives:
\[ (3 \cos \theta)^2 + (3 \sin \theta)^2 = 9 \cos^2 \theta + 9 \sin^2 \theta \]

Using the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ 9 (\cos^2 \theta + \sin^2 \theta) = 9 \cdot 1 = 9 \]

Thus, we have:
\[ x^2 + y^2 = 9 \]

### Conclusion
Since the coordinates of the centroid satisfy the equation of the circle, we conclude that the centroid of the triangle lies on the circle \(x^2 + y^2 = 9\).