A point moves so that the sum of squares of its distances from the points (1,2) and (-2,1) is always 6. then its locus is

Options

A. the straight line `y-(3)/(2)=-3(x+(1)/(2))`
B. the circle with centre `(-(1)/(2),(3)/(2))` and radius `(1)/(sqrt2)`
C. a parabola with focus (1,2) and directrix passing through (-2,1)
D. an ellipspe with foci (1,2) and (-2,1)

Answer

To find the locus of a point such that the sum of the squares of its distances from the points (1, 2) and (-2, 1) is always 6, we will follow these steps:

### Step 1: Define the point and distances
Let the moving point be \( P(h, k) \). The distances from \( P \) to the points \( A(1, 2) \) and \( B(-2, 1) \) can be calculated using the distance formula.

### Step 2: Calculate the distances
The distance \( AP \) from point \( A(1, 2) \) to point \( P(h, k) \) is given by:
\[ AP = \sqrt{(h - 1)^2 + (k - 2)^2} \]
The distance \( BP \) from point \( B(-2, 1) \) to point \( P(h, k) \) is given by:
\[ BP = \sqrt{(h + 2)^2 + (k - 1)^2} \]

### Step 3: Set up the equation
According to the problem, the sum of the squares of these distances is equal to 6:
\[ AP^2 + BP^2 = 6 \]
Substituting the expressions for \( AP \) and \( BP \):
\[ (h - 1)^2 + (k - 2)^2 + (h + 2)^2 + (k - 1)^2 = 6 \]

### Step 4: Expand the equation
Now, we will expand the squares:
\[ (h - 1)^2 = h^2 - 2h + 1 \]
\[ (k - 2)^2 = k^2 - 4k + 4 \]
\[ (h + 2)^2 = h^2 + 4h + 4 \]
\[ (k - 1)^2 = k^2 - 2k + 1 \]
Now substituting these into the equation:
\[ (h^2 - 2h + 1) + (k^2 - 4k + 4) + (h^2 + 4h + 4) + (k^2 - 2k + 1) = 6 \]

### Step 5: Combine like terms
Combining all the terms:
\[ 2h^2 + 2k^2 + (-2h + 4h - 4k - 2k) + (1 + 4 + 4 + 1) = 6 \]
This simplifies to:
\[ 2h^2 + 2k^2 + 2h - 6k + 10 = 6 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 2h^2 + 2k^2 + 2h - 6k + 4 = 0 \]
Dividing the entire equation by 2:
\[ h^2 + k^2 + h - 3k + 2 = 0 \]

### Step 7: Completing the square
To complete the square for \( h \) and \( k \):
1. For \( h^2 + h \):
\[ h^2 + h = (h + \frac{1}{2})^2 - \frac{1}{4} \]
2. For \( k^2 - 3k \):
\[ k^2 - 3k = (k - \frac{3}{2})^2 - \frac{9}{4} \]

Substituting these back into the equation:
\[ (h + \frac{1}{2})^2 - \frac{1}{4} + (k - \frac{3}{2})^2 - \frac{9}{4} + 2 = 0 \]
Combining constants:
\[ (h + \frac{1}{2})^2 + (k - \frac{3}{2})^2 - \frac{1}{4} - \frac{9}{4} + 2 = 0 \]
\[ (h + \frac{1}{2})^2 + (k - \frac{3}{2})^2 - \frac{10}{4} + 2 = 0 \]
\[ (h + \frac{1}{2})^2 + (k - \frac{3}{2})^2 = \frac{1}{2} \]

### Step 8: Final equation of the locus
This represents a circle with center at \( (-\frac{1}{2}, \frac{3}{2}) \) and radius \( \frac{1}{\sqrt{2}} \).