Let `P(alpha,beta)` be a point in the first quadrant. Circles are drawn through P touching the coordinate axes.
Equation of common chord of two circles is

Options

A. `x +y = alpha - beta`
B. `x +y = 2 sqrt(alpha beta)`
C. `x +y = alpha +beta`
D. `alpha^(2) -beta^(2) = 4 alpha beta`

Explanation

To solve the problem, we need to find the equation of the common chord of two circles that pass through the point \( P(\alpha, \beta) \) and touch the coordinate axes.

### Step-by-step Solution:

1. **Understanding the Circles**:
- The circles are drawn through the point \( P(\alpha, \beta) \) and touch the coordinate axes.
- Since the circles touch the axes, the center of each circle will be at \( (R, R) \), where \( R \) is the radius of the circle.

2. **Equation of the Circles**:
- The general equation of a circle with center \( (h, k) \) and radius \( R \) is given by:
\[ (x - h)^2 + (y - k)^2 = R^2 \]
- For our circles, substituting \( h = R \) and \( k = R \), we get:
\[ (x - R)^2 + (y - R)^2 = R^2 \]

3. **Expanding the Circle Equation**:
- Expanding the equation:
\[ (x - R)^2 + (y - R)^2 = R^2 \]
gives:
\[ x^2 - 2Rx + R^2 + y^2 - 2Ry + R^2 = R^2 \]
- Simplifying this, we have:
\[ x^2 + y^2 - 2Rx - 2Ry + R^2 = 0 \]
- This simplifies to:
\[ x^2 + y^2 - 2R(x + y) + R^2 = 0 \]

4. **Equation of the Common Chord**:
- For two circles with radii \( R_1 \) and \( R_2 \), the equation of the common chord can be expressed as:
\[ 2R_1R_2(x + y) = R_1^2 - R_2^2 \]
- Since both circles pass through the point \( P(\alpha, \beta) \), we can substitute \( x = \alpha \) and \( y = \beta \) into the equation:
\[ 2R_1R_2(\alpha + \beta) = R_1^2 - R_2^2 \]

5. **Finalizing the Equation**:
- Rearranging gives:
\[ R_1^2 - R_2^2 = 2R_1R_2(\alpha + \beta) \]
- This leads us to conclude that:
\[ \alpha + \beta = \frac{R_1 + R_2}{2} \]
- Therefore, the equation of the common chord simplifies to:
\[ x + y = \alpha + \beta \]

### Conclusion:
The equation of the common chord of the two circles is:
\[ x + y = \alpha + \beta \]