The axes being inclined at an angle ` omega ` find the centre and radius of the circle ` x^(2) + 2xy cos omega + y^(2) - 2gy - 2fy = 0`

Answer

To find the center and radius of the circle given by the equation:

\[ x^2 + 2xy \cos \omega + y^2 - 2gx - 2fy = 0 \]

we will follow these steps:

### Step 1: Rewrite the Circle Equation
We start by rewriting the given equation in a more standard form. The equation can be rearranged as:

\[ x^2 + 2xy \cos \omega + y^2 = 2gx + 2fy \]

### Step 2: Identify Coefficients
From the standard form of the circle equation, we can identify the coefficients:
- The term \(2xy \cos \omega\) indicates that the axes are inclined at an angle \(\omega\).
- The coefficients for \(x\) and \(y\) are \(-2g\) and \(-2f\) respectively.

### Step 3: Center of the Circle
The center \((h, k)\) of the circle can be derived from the equations:

1. \(h + k \cos \omega = g\)
2. \(k + h \cos \omega = f\)

We can solve these equations to find \(h\) and \(k\).

### Step 4: Solve for \(h\)
From the first equation, we can express \(k\) in terms of \(h\):

\[ k = \frac{g - h \cos \omega}{\cos \omega} \]

Substituting this into the second equation gives:

\[ \frac{g - h \cos \omega}{\cos \omega} + h \cos \omega = f \]

Multiplying through by \(\cos \omega\) to eliminate the denominator:

\[ g - h \cos \omega + h \cos^2 \omega = f \cos \omega \]

Rearranging gives:

\[ h(1 - \cos^2 \omega) = g - f \cos \omega \]

Thus, we find:

\[ h = \frac{g - f \cos \omega}{\sin^2 \omega} \]

### Step 5: Solve for \(k\)
Substituting \(h\) back into the equation for \(k\):

\[ k = \frac{g - \left(\frac{g - f \cos \omega}{\sin^2 \omega}\right) \cos \omega}{\cos \omega} \]

This simplifies to:

\[ k = \frac{f - g \cos \omega}{\sin^2 \omega} \]

### Step 6: Center Coordinates
Thus, the center of the circle \((h, k)\) is:

\[ \left( \frac{g - f \cos \omega}{\sin^2 \omega}, \frac{f - g \cos \omega}{\sin^2 \omega} \right) \]

### Step 7: Radius of the Circle
To find the radius \(r\), we use the standard form of the circle equation:

\[ h^2 + k^2 + 2hk \cos \omega - r^2 = 0 \]

Substituting \(h\) and \(k\) into this equation and solving for \(r^2\):

1. Calculate \(h^2\) and \(k^2\).
2. Calculate \(2hk \cos \omega\).
3. Set the equation equal to \(r^2\) and solve.

After simplification, we find:

\[ r = \sqrt{g^2 + f^2 - 2fg \cos \omega} \]

### Final Result
The center of the circle is:

\[ \left( \frac{g - f \cos \omega}{\sin^2 \omega}, \frac{f - g \cos \omega}{\sin^2 \omega} \right) \]

And the radius is:

\[ r = \sqrt{g^2 + f^2 - 2fg \cos \omega} \]