Find the locus of the point P such that `PA^2+PB^2=2c^2 ` :where A(a,0),B(-a,0) and `0 lt |a| lt |c|.`

Answer

To find the locus of the point \( P \) such that \( PA^2 + PB^2 = 2c^2 \), where \( A(a, 0) \) and \( B(-a, 0) \) and \( 0 < |a| < |c| \), we will follow these steps:

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

### Step 2: Calculate distances PA and PB
Using the distance formula, we can express \( PA \) and \( PB \):
- The distance \( PA \) from point \( P \) to point \( A(a, 0) \) is:
\[ PA = \sqrt{(x - a)^2 + (y - 0)^2} = \sqrt{(x - a)^2 + y^2} \]
- The distance \( PB \) from point \( P \) to point \( B(-a, 0) \) is:
\[ PB = \sqrt{(x + a)^2 + (y - 0)^2} = \sqrt{(x + a)^2 + y^2} \]

### Step 3: Square the distances
Now we square both distances:
- \( PA^2 = (x - a)^2 + y^2 \)
- \( PB^2 = (x + a)^2 + y^2 \)

### Step 4: Set up the equation
According to the problem, we have:
\[ PA^2 + PB^2 = 2c^2 \]
Substituting the squared distances:
\[ (x - a)^2 + y^2 + (x + a)^2 + y^2 = 2c^2 \]

### Step 5: Simplify the equation
Now, simplify the left-hand side:
\[ (x - a)^2 + (x + a)^2 + 2y^2 = 2c^2 \]
Expanding the squares:
\[ (x^2 - 2ax + a^2) + (x^2 + 2ax + a^2) + 2y^2 = 2c^2 \]
Combining like terms:
\[ 2x^2 + 2a^2 + 2y^2 = 2c^2 \]

### Step 6: Divide through by 2
Dividing the entire equation by 2 gives:
\[ x^2 + y^2 + a^2 = c^2 \]

### Step 7: Rearranging the equation
Rearranging this, we get:
\[ x^2 + y^2 = c^2 - a^2 \]

### Conclusion
This is the equation of a circle centered at the origin with radius \( \sqrt{c^2 - a^2} \).