Question

Let A be the point (1, 2) and B be any point on the curve \( x^2 + y^2 = 16 \). If the centre of the locus of the point P, which divides the line segment AB in the ratio 3:2 is the point C (\( \alpha, \beta \)), then the length of the line segment AC is:

• A. \( \frac{6 \sqrt{5}}{5} \)
• B. \( \frac{2 \sqrt{5}}{5} \)
• C. \( \frac{3 \sqrt{5}}{5} \)
• D. \( \frac{4 \sqrt{5}}{5} \)

Answer 1

The Correct Option is C

Step 1: Write the coordinates of points A and B Let \( A(1, 2) \) and the coordinates of point \( B \) be \( B(4 \cos \theta, 4 \sin \theta) \) because \( x^2 + y^2 = 16 \) represents a circle of radius 4. 
Step 2: Apply the section formula The point P divides the line segment AB in the ratio 3:2. Using the section formula: \[ \left( \frac{3x_B + 2x_A}{5}, \frac{3y_B + 2y_A}{5} \right) \] Substitute the coordinates of \( A(1, 2) \) and \( B(4 \cos \theta, 4 \sin \theta) \) into the formula: \[ P = \left( \frac{3(4 \cos \theta) + 2(1)}{5}, \frac{3(4 \sin \theta) + 2(2)}{5} \right) \] Simplifying, we get: \[ P = \left( \frac{12 \cos \theta + 2}{5}, \frac{12 \sin \theta + 4}{5} \right) \] 
Step 3: Find the coordinates of the centre of the locus of P The centre of the locus of the point P is the midpoint of the line segment AB. The midpoint is given by: \[ \left( \frac{1 + 4 \cos \theta}{2}, \frac{2 + 4 \sin \theta}{2} \right) \] This is the point \( C (\alpha, \beta) \). 
Step 4: Calculate the length of the line segment AC Using the distance formula between \( A(1, 2) \) and \( C(\alpha, \beta) \), we get: \[ AC = \sqrt{\left( 1 - \frac{2}{5} \right)^2 + \left( 2 - \frac{4}{5} \right)^2} \] Simplifying: \[ AC = \sqrt{\left( \frac{3}{5} \right)^2 + \left( \frac{6}{5} \right)^2} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \frac{3 \sqrt{5}}{5} \] Thus, the length of the line segment AC is \( \frac{3 \sqrt{5}}{5} \).