The axes being inclined at `60^(@)` find the equation to the circle whose centre is the point ( - 3, - 5) and whose radius is 6.

Answer

To find the equation of the circle whose center is at the point \((-3, -5)\) and whose radius is \(6\), with the axes inclined at \(60^\circ\), we will use the general form of the equation of a circle in inclined coordinates.

### Step-by-Step Solution:

1. **Identify the center and radius**:
- The center of the circle is given as \((-3, -5)\), so \(h = -3\) and \(k = -5\).
- The radius \(r = 6\).

2. **Determine the angle**:
- The angle of inclination of the axes is given as \(\omega = 60^\circ\).

3. **Use the standard equation of the circle in inclined coordinates**:
The standard equation of a circle when the axes are inclined at an angle \(\omega\) is:
\[ x^2 + y^2 + 2xy \cos(\omega) - 2hx - 2ky + (h^2 + k^2 + 2hk \cos(\omega) - r^2) = 0 \]

4. **Substitute the values**:
- Calculate \(\cos(60^\circ) = \frac{1}{2}\).
- Substitute \(h = -3\), \(k = -5\), and \(r = 6\) into the equation:
\[ x^2 + y^2 + 2xy \cdot \frac{1}{2} - 2(-3)x - 2(-5)y + \left((-3)^2 + (-5)^2 + 2(-3)(-5) \cdot \frac{1}{2} - 6^2\right) = 0 \]

5. **Simplify the equation**:
- The equation becomes:
\[ x^2 + y^2 + xy + 6x + 10y + \left(9 + 25 + 15 - 36\right) = 0 \]
- Calculate the constant term:
\[ 9 + 25 + 15 - 36 = 13 \]
- Thus, the equation simplifies to:
\[ x^2 + y^2 + xy + 6x + 10y + 13 = 0 \]

### Final Equation:
The equation of the circle is:
\[ x^2 + y^2 + xy + 6x + 10y + 13 = 0 \]