The axes being inclined at `45^(@)` find the equation to the circle whose centre is the point (2, 3) and whose radius is 4.

Answer

To find the equation of a circle whose center is at the point (2, 3) and radius is 4, with the axes inclined at 45 degrees, we can follow these steps:

### Step 1: Write the general equation of a circle
The general equation of a circle in a coordinate system with axes inclined at an angle \( \omega \) is given by:

\[ x^2 + y^2 + 2xy \cos(\omega) - 2hx - 2ky + (h^2 + k^2 + 2hk \cos(\omega) - r^2) = 0 \]

### Step 2: Identify the parameters
Here, we have:
- Center \( (h, k) = (2, 3) \)
- Radius \( r = 4 \)
- Angle of inclination \( \omega = 45^\circ \)

### Step 3: Substitute the values
We know that \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \). Now we can substitute the values into the equation:

\[ x^2 + y^2 + 2xy \cdot \frac{1}{\sqrt{2}} - 2 \cdot 2x - 2 \cdot 3y + \left(2^2 + 3^2 + 2 \cdot 2 \cdot 3 \cdot \frac{1}{\sqrt{2}} - 4^2\right) = 0 \]

### Step 4: Simplify the equation
Calculating the constant term:

1. Calculate \( 2^2 + 3^2 = 4 + 9 = 13 \)
2. Calculate \( 2 \cdot 2 \cdot 3 \cdot \frac{1}{\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \)
3. Calculate \( 4^2 = 16 \)

Now substituting these values into the constant term:

\[ 13 + 6\sqrt{2} - 16 = -3 + 6\sqrt{2} \]

### Step 5: Write the complete equation
Now we can write the complete equation:

\[ x^2 + y^2 + \sqrt{2}xy - 4x - 6y + (6\sqrt{2} - 3) = 0 \]

### Final Equation
Thus, the final equation of the circle is:

\[ x^2 + y^2 + \sqrt{2}xy - 4x - 6y + (6\sqrt{2} - 3) = 0 \]