The axes being inclined at an angle ` omega ` find the equation to the circle whose diameter is the straight line joining the points
(x'y') and (x'', y'')

Answer

To find the equation of the circle whose diameter is the straight line joining the points \((x', y')\) and \((x'', y'')\) with the axes inclined at an angle \(\omega\), we can follow these steps:

### Step 1: Identify the center and radius of the circle
The center \((h, k)\) of the circle can be calculated as the midpoint of the endpoints of the diameter. Thus, we have:
\[ h = \frac{x' + x''}{2}, \quad k = \frac{y' + y''}{2} \]

### Step 2: Calculate the radius of the circle
The radius \(r\) of the circle can be found using the distance formula between the two endpoints of the diameter. The distance \(d\) between the points \((x', y')\) and \((x'', y'')\) is given by:
\[ d = \sqrt{(x' - x'')^2 + (y' - y'')^2} \]
Since \(d\) is the diameter, the radius \(r\) is:
\[ r = \frac{d}{2} = \frac{1}{2} \sqrt{(x' - x'')^2 + (y' - y'')^2} \]

### Step 3: Write the general equation of the circle
The general equation of a circle in an inclined coordinate system can be expressed as:
\[ x^2 + y^2 + 2xy \cos \omega - 2hx - 2ky + (h^2 + k^2 + 2hk \cos \omega - r^2) = 0 \]

### Step 4: Substitute \(h\), \(k\), and \(r\) into the equation
Substituting \(h\), \(k\), and \(r\) into the general equation gives:
\[ x^2 + y^2 + 2xy \cos \omega - 2\left(\frac{x' + x''}{2}\right)x - 2\left(\frac{y' + y''}{2}\right)y + \left(\left(\frac{x' + x''}{2}\right)^2 + \left(\frac{y' + y''}{2}\right)^2 + 2\left(\frac{x' + x''}{2}\right)\left(\frac{y' + y''}{2}\right) \cos \omega - \left(\frac{1}{2} \sqrt{(x' - x'')^2 + (y' - y'')^2}\right)^2\right) = 0 \]

### Step 5: Simplify the equation
After substituting and simplifying, we can rearrange the terms to obtain the final equation of the circle.

### Final Equation
The final equation of the circle can be expressed as:
\[ x^2 + y^2 + 2xy \cos \omega - (x' + x'')y \cos \omega - (y' + y'')x \cos \omega + \text{constant} = 0 \]