(i) Find the equation a circle passing through the point `(2+3costheta,1+3sintheta)` where `'theta'` is a parameter.
(ii) Prove that the equations `x=acostheta+bsintheta and y=asintheta-bcostheta` represents a circle.

Answer

Let's solve the question step by step.

### Part (i): Find the equation of a circle passing through the point \((2 + 3 \cos \theta, 1 + 3 \sin \theta)\).

1. **Identify the coordinates**:
The point given is \((x, y) = (2 + 3 \cos \theta, 1 + 3 \sin \theta)\).

2. **Express \(\cos \theta\) and \(\sin \theta\)**:
From the x-coordinate:
\[ \cos \theta = \frac{x - 2}{3} \]
From the y-coordinate:
\[ \sin \theta = \frac{y - 1}{3} \]

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

4. **Simplify the equation**:
Multiply through by \(9\) (which is \(3^2\)):
\[ (x - 2)^2 + (y - 1)^2 = 9 \]

5. **Identify the center and radius**:
The equation \((x - 2)^2 + (y - 1)^2 = 9\) represents a circle with:
- Center: \((2, 1)\)
- Radius: \(3\) (since \(r^2 = 9\))

### Part (ii): Prove that the equations \(x = a \cos \theta + b \sin \theta\) and \(y = a \sin \theta - b \cos \theta\) represent a circle.

1. **Express \(x\) and \(y\)**:
Given:
\[ x = a \cos \theta + b \sin \theta \]
\[ y = a \sin \theta - b \cos \theta \]

2. **Square both equations**:
\[ x^2 = (a \cos \theta + b \sin \theta)^2 = a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta \]
\[ y^2 = (a \sin \theta - b \cos \theta)^2 = a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta \]

3. **Add the two equations**:
\[ x^2 + y^2 = (a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta) \]

4. **Combine like terms**:
\[ x^2 + y^2 = a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) + (2ab - 2ab) \cos \theta \sin \theta \]
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ x^2 + y^2 = a^2 + b^2 \]

5. **Final equation**:
The equation \(x^2 + y^2 = a^2 + b^2\) represents a circle centered at the origin \((0, 0)\) with radius \(\sqrt{a^2 + b^2}\).

### Summary of Solutions:
- **Part (i)**: The equation of the circle is \((x - 2)^2 + (y - 1)^2 = 9\).
- **Part (ii)**: The equations represent a circle with the equation \(x^2 + y^2 = a^2 + b^2\).