The line joining `(5,0)` to `(10cos theta,10sin theta)` is divided internally in the ratio `2:3` at P then the locus of P is

Options

A. `4pi" units"`
B. `16pi" units"`
C. `16pi" units"`
D. `6pi" units"`

Answer

To find the locus of point P that divides the line segment joining the points \( A(5, 0) \) and \( B(10 \cos \theta, 10 \sin \theta) \) in the ratio \( 2:3 \), we can use the section formula.

### Step-by-Step Solution:

1. **Identify the Points**:
- Let \( A(5, 0) \) and \( B(10 \cos \theta, 10 \sin \theta) \).

2. **Apply the Section Formula**:
The coordinates of point P that divides the line segment joining points \( A \) and \( B \) in the ratio \( m:n \) is given by:
\[ P\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]
Here, \( m = 2 \), \( n = 3 \), \( x_1 = 5 \), \( y_1 = 0 \), \( x_2 = 10 \cos \theta \), and \( y_2 = 10 \sin \theta \).

3. **Calculate the x-coordinate of P**:
\[ h = \frac{2(10 \cos \theta) + 3(5)}{2 + 3} = \frac{20 \cos \theta + 15}{5} = 4 \cos \theta + 3 \]

4. **Calculate the y-coordinate of P**:
\[ k = \frac{2(10 \sin \theta) + 3(0)}{2 + 3} = \frac{20 \sin \theta}{5} = 4 \sin \theta \]

5. **Express cos θ and sin θ in terms of h and k**:
From the equations derived:
- \( \cos \theta = \frac{h - 3}{4} \)
- \( \sin \theta = \frac{k}{4} \)

6. **Use the Pythagorean Identity**:
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). Substituting the expressions for \( \sin \theta \) and \( \cos \theta \):
\[ \left( \frac{k}{4} \right)^2 + \left( \frac{h - 3}{4} \right)^2 = 1 \]

7. **Simplify the Equation**:
\[ \frac{k^2}{16} + \frac{(h - 3)^2}{16} = 1 \]
Multiply through by 16:
\[ k^2 + (h - 3)^2 = 16 \]

8. **Rearranging the Equation**:
Expanding \( (h - 3)^2 \):
\[ k^2 + (h^2 - 6h + 9) = 16 \]
Therefore:
\[ h^2 + k^2 - 6h + 9 = 16 \]
Simplifying gives:
\[ h^2 + k^2 - 6h - 7 = 0 \]

9. **Final Form**:
Replacing \( h \) with \( x \) and \( k \) with \( y \):
\[ x^2 + y^2 - 6x - 7 = 0 \]

### Conclusion:
The locus of point P is given by the equation:
\[ x^2 + y^2 - 6x - 7 = 0 \]