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

Updated On: Jan 11, 2025
  • \(\frac{\sqrt{19}}{7}\lambda\)
  • \(\frac{\sqrt{19}}{5}\lambda\)
  • \(\frac{{2}}{3}\lambda\)
  • \(\frac{{3}}{5}\lambda\)
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The Correct Option is B

Solution and Explanation

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

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