A(2, 0) and B(4, 0) are two given points. A point P moves so that `PA^(2)+PB^(2)=10`. Find the locus of P.

Answer

To find the locus of point P that moves such that \( PA^2 + PB^2 = 10 \), where \( A(2, 0) \) and \( B(4, 0) \), we can follow these steps:

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

### Step 2: Calculate distances PA and PB
Using the distance formula, we can express the distances \( PA \) and \( PB \):
- The distance \( PA \) from point P to point A is:
\[ PA = \sqrt{(h - 2)^2 + (k - 0)^2} = \sqrt{(h - 2)^2 + k^2} \]
- The distance \( PB \) from point P to point B is:
\[ PB = \sqrt{(h - 4)^2 + (k - 0)^2} = \sqrt{(h - 4)^2 + k^2} \]

### Step 3: Set up the equation based on the given condition
According to the problem, we have:
\[ PA^2 + PB^2 = 10 \]
Substituting the expressions for \( PA \) and \( PB \):
\[ (h - 2)^2 + k^2 + (h - 4)^2 + k^2 = 10 \]

### Step 4: Simplify the equation
Combine the terms:
\[ (h - 2)^2 + (h - 4)^2 + 2k^2 = 10 \]
Expanding the squares:
\[ (h^2 - 4h + 4) + (h^2 - 8h + 16) + 2k^2 = 10 \]
Combine like terms:
\[ 2h^2 - 12h + 20 + 2k^2 = 10 \]
Subtract 10 from both sides:
\[ 2h^2 - 12h + 2k^2 + 10 = 0 \]

### Step 5: Divide the entire equation by 2
To simplify, divide the equation by 2:
\[ h^2 - 6h + k^2 + 5 = 0 \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ h^2 + k^2 - 6h + 5 = 0 \]

### Step 7: Complete the square for the \( h \) terms
To complete the square for \( h \):
\[ (h^2 - 6h + 9) + k^2 - 9 + 5 = 0 \]
This simplifies to:
\[ (h - 3)^2 + k^2 - 4 = 0 \]
Thus, we have:
\[ (h - 3)^2 + k^2 = 4 \]

### Step 8: Identify the locus
This is the equation of a circle with center at \( (3, 0) \) and radius \( 2 \).

### Final Result
The locus of point P is:
\[ (h - 3)^2 + k^2 = 4 \]