A and B are two fixed points and P moves so that PA=nPB. Show that locus of P is a circle and for different values of n all the circles have a common radical axis.

Answer

To solve the problem, we need to show that the locus of point P, which moves such that \( PA = n \cdot PB \), is a circle, and that for different values of \( n \), all such circles share a common radical axis.

### Step 1: Define the Points
Let \( A \) and \( B \) be two fixed points in the Cartesian plane with coordinates:
- \( A(x_1, y_1) \)
- \( B(x_2, y_2) \)

Let \( P(x, y) \) be the moving point.

### Step 2: Express Distances
The distances from point P to points A and B can be expressed as:
- \( PA = \sqrt{(x - x_1)^2 + (y - y_1)^2} \)
- \( PB = \sqrt{(x - x_2)^2 + (y - y_2)^2} \)

### Step 3: Set Up the Equation
Given that \( PA = n \cdot PB \), we can write:
\[ \sqrt{(x - x_1)^2 + (y - y_1)^2} = n \cdot \sqrt{(x - x_2)^2 + (y - y_2)^2} \]

### Step 4: Square Both Sides
To eliminate the square roots, we square both sides:
\[ (x - x_1)^2 + (y - y_1)^2 = n^2 \cdot \left((x - x_2)^2 + (y - y_2)^2\right) \]

### Step 5: Expand Both Sides
Expanding both sides gives:
\[ (x - x_1)^2 + (y - y_1)^2 = n^2 \left((x - x_2)^2 + (y - y_2)^2\right) \]
\[ x^2 - 2xx_1 + x_1^2 + y^2 - 2yy_1 + y_1^2 = n^2 \left(x^2 - 2xx_2 + x_2^2 + y^2 - 2yy_2 + y_2^2\right) \]

### Step 6: Rearrange the Equation
Rearranging the equation, we get:
\[ x^2 + y^2 - 2xx_1 - 2yy_1 + x_1^2 + y_1^2 = n^2 \left(x^2 + y^2 - 2xx_2 - 2yy_2 + x_2^2 + y_2^2\right) \]

### Step 7: Collect Like Terms
Bringing all terms involving \( x \) and \( y \) to one side results in:
\[ (1 - n^2)x^2 + (1 - n^2)y^2 - 2x(x_1 - n^2x_2) - 2y(y_1 - n^2y_2) + (x_1^2 + y_1^2 - n^2(x_2^2 + y_2^2)) = 0 \]

### Step 8: Identify the Circle Equation
This equation can be rearranged into the standard form of a circle:
\[ Ax^2 + Ay^2 + 2Gx + 2Fy + C = 0 \]
where \( A = 1 - n^2 \), \( G = -(x_1 - n^2x_2) \), \( F = -(y_1 - n^2y_2) \), and \( C = x_1^2 + y_1^2 - n^2(x_2^2 + y_2^2) \).

### Step 9: Show Common Radical Axis
For two different values \( n_1 \) and \( n_2 \), the equations of the circles can be derived similarly. Subtracting the two equations will eliminate the quadratic terms, leading to a linear equation in \( x \) and \( y \), which represents the radical axis.

### Conclusion
Thus, we have shown that the locus of point P is a circle and that for different values of \( n \), all these circles share a common radical axis.