Find the locus of P If the distance of P from (3,0) is twice the distance of P from (-3,0)

Answer

To find the locus of point P such that the distance from P to (3,0) is twice the distance from P to (-3,0), we can follow these steps:

### Step 1: Define the points and the point P
Let the point P be represented by the coordinates (x, y). We have two fixed points:
- A = (3, 0)
- B = (-3, 0)

### Step 2: Write the distance formulas
The distance from P to A (3, 0) is given by:
\[ PA = \sqrt{(x - 3)^2 + (y - 0)^2} = \sqrt{(x - 3)^2 + y^2} \]

The distance from P to B (-3, 0) is given by:
\[ PB = \sqrt{(x + 3)^2 + (y - 0)^2} = \sqrt{(x + 3)^2 + y^2} \]

### Step 3: Set up the equation based on the given condition
According to the problem, the distance from P to A is twice the distance from P to B:
\[ PA = 2 \cdot PB \]

Substituting the distance formulas, we have:
\[ \sqrt{(x - 3)^2 + y^2} = 2 \cdot \sqrt{(x + 3)^2 + y^2} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ (x - 3)^2 + y^2 = 4 \cdot ((x + 3)^2 + y^2) \]

### Step 5: Expand both sides
Expanding the left side:
\[ (x - 3)^2 + y^2 = x^2 - 6x + 9 + y^2 \]

Expanding the right side:
\[ 4 \cdot ((x + 3)^2 + y^2) = 4 \cdot (x^2 + 6x + 9 + y^2) = 4x^2 + 24x + 36 + 4y^2 \]

### Step 6: Set the equation
Now we set the expanded forms equal to each other:
\[ x^2 - 6x + 9 + y^2 = 4x^2 + 24x + 36 + 4y^2 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ x^2 - 4x^2 - 6x - 24x + 9 - 36 + y^2 - 4y^2 = 0 \]
This simplifies to:
\[ -3x^2 - 30x - 3y^2 - 27 = 0 \]

### Step 8: Divide through by -3
Dividing the entire equation by -3 yields:
\[ x^2 + 10x + y^2 + 9 = 0 \]

### Step 9: Complete the square
To complete the square for the x terms:
\[ (x^2 + 10x + 25) + y^2 = 16 \]
This can be rewritten as:
\[ (x + 5)^2 + y^2 = 16 \]

### Conclusion
The locus of point P is a circle centered at (-5, 0) with a radius of 4.

---