Question

A line segment AB of length λ moves such that the points A and B remain on the periphery of a circle of radius λ. Then the locus of the point, that divides the line segment AB in the ratio 2 : 3, is a circle of radius

• A. \(\frac{\sqrt{19}}{7}\lambda\)
• B. \(\frac{\sqrt{19}}{5}\lambda\)
• C. \(\frac{{2}}{3}\lambda\)
• D. \(\frac{{3}}{5}\lambda\)

Answer 1

The Correct Option is B

Given: The triangle \( OAB \) is an equilateral triangle, and the point \( P \) divides \( AB \) in the ratio \( 2:3 \).

• Step 1: Geometry of the equilateral triangle:
  • \( OAP = 60^\circ \)
  • \( AP = \frac{2\lambda}{5} \)
  • \( OA = OB = \lambda \)
• Step 2: Using the cosine rule:

\( \cos 60^\circ = \frac{OA^2 + AP^2 - OP^2}{2 \cdot OA \cdot AP} \).

• Substitute \( \cos 60^\circ = \frac{1}{2} \):

\( \frac{1}{2} = \frac{\lambda^2 + \frac{4\lambda^2}{25} - OP^2}{2 \cdot \lambda \cdot \frac{2\lambda}{5}} \).

• Simplify the denominator:

\( \frac{1}{2} = \frac{\lambda^2 + \frac{4\lambda^2}{25} - OP^2}{\frac{4\lambda^2}{5}} \).

• Multiply through by \( \frac{4\lambda^2}{5} \):

\( \frac{2\lambda^2}{5} = \lambda^2 + \frac{4\lambda^2}{25} - OP^2 \).

• Step 3: Simplify for \( OP^2 \):

\( OP^2 = \lambda^2 + \frac{4\lambda^2}{25} - \frac{2\lambda^2}{5} \).

• Express all terms with a common denominator:

\( OP^2 = \frac{25\lambda^2}{25} + \frac{4\lambda^2}{25} - \frac{10\lambda^2}{25} \).

• Combine:

\( OP^2 = \frac{19\lambda^2}{25} \).

• Step 4: Solve for \( OP \):

\( OP = \sqrt{\frac{19\lambda^2}{25}} = \frac{\sqrt{19}}{5} \lambda \).

Final Answer: \( OP = \frac{\sqrt{19}}{5} \lambda \).