The locus of a point P which divides the line joining (1, 0) and `(2 cos theta, 2sin theta)` internally in the ratio 2 : 3 for all `theta` is

Options

A. a straight line
B. a circle
C. a pair of straight lines
D. a parabola

Answer

To find the locus of the point P that divides the line segment joining the points (1, 0) and (2 cos θ, 2 sin θ) internally in the ratio 2:3, we can follow these steps:

### Step 1: Identify the Points and the Ratio
Let A = (1, 0) and B = (2 cos θ, 2 sin θ). The point P divides the line segment AB in the ratio 2:3.

### Step 2: Use the Section Formula
According to the section formula, if a point P divides the line segment joining points A(x1, y1) and B(x2, y2) in the ratio m:n, the coordinates of P (x, y) can be given by:
\[ x = \frac{m x_2 + n x_1}{m + n} \]
\[ y = \frac{m y_2 + n y_1}{m + n} \]

### Step 3: Substitute the Values
Here, m = 2, n = 3, \(x_1 = 1\), \(y_1 = 0\), \(x_2 = 2 \cos \theta\), and \(y_2 = 2 \sin \theta\).

Substituting into the formula for x:
\[ x = \frac{2(2 \cos \theta) + 3(1)}{2 + 3} = \frac{4 \cos \theta + 3}{5} \]

Now for y:
\[ y = \frac{2(2 \sin \theta) + 3(0)}{2 + 3} = \frac{4 \sin \theta}{5} \]

### Step 4: Express cos θ and sin θ in terms of x and y
From the equations derived:
1. \(4 \cos \theta + 3 = 5x\) ⇒ \(4 \cos \theta = 5x - 3\) ⇒ \(\cos \theta = \frac{5x - 3}{4}\) (Equation 1)
2. \(4 \sin \theta = 5y\) ⇒ \(\sin \theta = \frac{5y}{4}\) (Equation 2)

### Step 5: Use the Pythagorean Identity
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \left(\frac{5y}{4}\right)^2 + \left(\frac{5x - 3}{4}\right)^2 = 1 \]

### Step 6: Simplify the Equation
Expanding both terms:
\[ \frac{25y^2}{16} + \frac{(5x - 3)^2}{16} = 1 \]

Multiply through by 16 to eliminate the fraction:
\[ 25y^2 + (5x - 3)^2 = 16 \]

### Step 7: Expand and Rearrange
Expanding \((5x - 3)^2\):
\[ 25y^2 + (25x^2 - 30x + 9) = 16 \]
\[ 25x^2 + 25y^2 - 30x + 9 - 16 = 0 \]
\[ 25x^2 + 25y^2 - 30x - 7 = 0 \]

### Step 8: Final Form
This equation can be rearranged to represent the locus of point P:
\[ 25x^2 + 25y^2 - 30x + 7 = 0 \]

### Conclusion
The locus of the point P is a circle.