Find the coordinates of the centre of the circle
`r = A cos theta + B sin theta `

Answer

To find the coordinates of the center of the circle given by the equation \( r = A \cos \theta + B \sin \theta \), we will follow these steps:

### Step 1: Rewrite the equation in a standard form
The given equation is in polar coordinates. We need to convert it into a form that allows us to identify the center and radius of the circle. The standard form of a circle in polar coordinates is:
\[ r^2 - 2r r_0 \cos(\theta - \alpha) = r_0^2 \]
where \( (r_0, \alpha) \) are the polar coordinates of the center.

### Step 2: Multiply both sides by \( r \)
To manipulate the equation, we multiply both sides by \( r \):
\[ r^2 = A r \cos \theta + B r \sin \theta \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ r^2 - A r \cos \theta - B r \sin \theta = 0 \]

### Step 4: Identify \( r_0 \) and \( \alpha \)
To match the standard form, we need to express the equation in terms of \( r_0 \) and \( \alpha \). We can set:
\[ r_0 = \frac{1}{2} \sqrt{A^2 + B^2} \]
and we can find \( \alpha \) using:
\[ \cos \alpha = \frac{A}{\sqrt{A^2 + B^2}}, \quad \sin \alpha = \frac{B}{\sqrt{A^2 + B^2}} \]
Thus, \( \tan \alpha = \frac{B}{A} \).

### Step 5: Write the coordinates of the center
The coordinates of the center of the circle in polar coordinates are:
\[ (r_0, \alpha) = \left(\frac{1}{2} \sqrt{A^2 + B^2}, \tan^{-1} \left(\frac{B}{A}\right)\right) \]

### Final Answer
The coordinates of the center of the circle are:
\[ \left(\frac{1}{2} \sqrt{A^2 + B^2}, \tan^{-1} \left(\frac{B}{A}\right)\right) \]