The locus of a point which divides the join of A(-1,1) and a variable point P on the circle `x^(2)+y^(2)=4` in the ratio 3:2 is

Options

A. `25(x^(2)+y^(2))+20(x+y)+28=0`
B. `25(x^(2)+y^(2))-20(x+y)+28=0`
C. `25(x^(2)+y^(2))+20(x-y)+28=0`
D. `25(x^(2)+y^(2))+20(x-y)-28=0`

Answer

To find the locus of a point that divides the line segment joining point A(-1, 1) and a variable point P on the circle \(x^2 + y^2 = 4\) in the ratio 3:2, we can follow these steps:

### Step 1: Parametrize the Point P
The equation of the circle \(x^2 + y^2 = 4\) can be parametrized as:
\[ P(x, y) = (2 \cos \theta, 2 \sin \theta) \]
where \(\theta\) is a variable angle.

### Step 2: Use the Section Formula
Let the point that divides the line segment joining A and P in the ratio 3:2 be denoted as \(L(h, k)\). According to the section formula, the coordinates of point L can be calculated as:
\[ h = \frac{3 \cdot x_P + 2 \cdot x_A}{3 + 2} = \frac{3(2 \cos \theta) + 2(-1)}{5} = \frac{6 \cos \theta - 2}{5} \]
\[ k = \frac{3 \cdot y_P + 2 \cdot y_A}{3 + 2} = \frac{3(2 \sin \theta) + 2(1)}{5} = \frac{6 \sin \theta + 2}{5} \]

### Step 3: Express Cosine and Sine in Terms of h and k
From the equations for h and k, we can express \(\cos \theta\) and \(\sin \theta\):
\[ 6 \cos \theta = 5h + 2 \implies \cos \theta = \frac{5h + 2}{6} \]
\[ 6 \sin \theta = 5k - 2 \implies \sin \theta = \frac{5k - 2}{6} \]

### Step 4: Use the Pythagorean Identity
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\), we substitute the expressions for \(\cos \theta\) and \(\sin \theta\):
\[ \left(\frac{5h + 2}{6}\right)^2 + \left(\frac{5k - 2}{6}\right)^2 = 1 \]

### Step 5: Simplify the Equation
Squaring both terms:
\[ \frac{(5h + 2)^2}{36} + \frac{(5k - 2)^2}{36} = 1 \]
Multiplying through by 36 to eliminate the denominator:
\[ (5h + 2)^2 + (5k - 2)^2 = 36 \]

### Step 6: Expand and Rearrange
Expanding both squares:
\[ (25h^2 + 20h + 4) + (25k^2 - 20k + 4) = 36 \]
Combining like terms:
\[ 25h^2 + 25k^2 + 20h - 20k + 8 = 36 \]
Rearranging gives:
\[ 25h^2 + 25k^2 + 20h - 20k - 28 = 0 \]

### Final Step: Replace h and k with x and y
Since \(h\) and \(k\) represent the coordinates of the locus point, we replace them with \(x\) and \(y\):
\[ 25x^2 + 25y^2 + 20x - 20y - 28 = 0 \]

### Conclusion
The locus of the point that divides the line segment in the ratio 3:2 is given by:
\[ 25x^2 + 25y^2 + 20x - 20y - 28 = 0 \]