The locus of P for which the distance from P to origin is double the distance from P to ` ( 1, 2 ) ` is

Answer

To find the locus of point \( P \) such that the distance from \( P \) to the origin is double the distance from \( P \) to the point \( (1, 2) \), we can follow these steps:

### Step 1: Define the point \( P \)
Let the coordinates of point \( P \) be \( (x, y) \).

### Step 2: Write the distance from \( P \) to the origin
The distance from point \( P \) to the origin \( (0, 0) \) is given by the formula:
\[ PO = \sqrt{x^2 + y^2} \]

### Step 3: Write the distance from \( P \) to the point \( (1, 2) \)
The distance from point \( P \) to the point \( A(1, 2) \) is given by:
\[ PA = \sqrt{(x - 1)^2 + (y - 2)^2} \]

### Step 4: Set up the equation based on the given condition
According to the problem, the distance from \( P \) to the origin is double the distance from \( P \) to \( (1, 2) \). Therefore, we can write:
\[ \sqrt{x^2 + y^2} = 2 \cdot \sqrt{(x - 1)^2 + (y - 2)^2} \]

### Step 5: Square both sides to eliminate the square roots
Squaring both sides gives us:
\[ x^2 + y^2 = 4 \cdot ((x - 1)^2 + (y - 2)^2) \]

### Step 6: Expand the right-hand side
Expanding the right-hand side:
\[ x^2 + y^2 = 4 \cdot ((x^2 - 2x + 1) + (y^2 - 4y + 4)) \]
\[ = 4(x^2 - 2x + 1 + y^2 - 4y + 4) \]
\[ = 4(x^2 + y^2 - 2x - 4y + 5) \]
\[ = 4x^2 + 4y^2 - 8x - 16y + 20 \]

### Step 7: Rearrange the equation
Now, we equate both sides:
\[ x^2 + y^2 = 4x^2 + 4y^2 - 8x - 16y + 20 \]
Rearranging gives:
\[ 0 = 4x^2 + 4y^2 - x^2 - y^2 - 8x - 16y + 20 \]
\[ 0 = 3x^2 + 3y^2 - 8x - 16y + 20 \]

### Step 8: Final equation of the locus
Thus, the locus of point \( P \) is given by:
\[ 3x^2 + 3y^2 - 8x - 16y + 20 = 0 \]