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

Given: The triangle OAB is an equilateral triangle, and the point P divides AB in the ratio 2:3. Find: The radius of the locus of point P. From the given working: - OAP = 60^ - AP = 2 5 - OA = OB = Using the cosine rule in triangle OAP, 60^ = OA^2 + AP^2 - OP^22 OA AP Substituting the given values, 12 = ^2 + 4 ^225 - OP^22 2 5 So, 12 = ^2 + 4 ^225 - OP^2 4 ^25 Multiplying through by 4 ^25, 2 ^25 = ^2 + 4 ^225 - OP^2 Hence, OP^2 = ^2 + 4 ^225 - 2 ^25 Writing all terms with denominator 25, OP^2 = 25 ^225 + 4 ^225 - 10 ^225 Therefore, OP^2 = 19 ^225 and so, OP = 19 ^225 = 195 Therefore, the radius of the locus is 195. The solution concludes this value, although it labels the correct option as B. Comparing with the given options, 19 5 corresponds to option D. Hence the correct option should be D.

A line segment AB of length moves such that the points A and B re… — Solution | JEE Mathematics

MCQMediumJEE 2023Circle Equation & Properties

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

A
$\frac{2\lambda}{3}$
B
$\frac{\sqrt{19}\lambda}{7}$
C
$\frac{3\lambda}{5}$
D
$\frac{\sqrt{19}\lambda}{5}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

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

Find: The radius of the locus of point $P$ .

From the given working:

$\angle OAP = 60^\circ$
$AP = \frac{2\lambda}{5}$
$OA = OB = \lambda$

Using the cosine rule in triangle $OAP$ ,

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

Substituting the given values,

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

So,

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

Multiplying through by $\frac{4\lambda^2}{5}$ ,

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

Hence,

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

Writing all terms with denominator $25$ ,

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

Therefore,

OP^2 = \frac{19\lambda^2}{25}

and so,

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

Therefore, the radius of the locus is $\frac{\sqrt{19}}{5}\lambda$ .

The solution concludes this value, although it labels the correct option as B. Comparing with the given options, $\frac{\sqrt{19}\lambda}{5}$ corresponds to option D. Hence the correct option should be D.

Common mistakes

Using the ratio $2:3$ incorrectly by taking $AP = \frac{2\lambda}{3}$ . This is wrong because the whole segment $AB$ is divided into $2+3=5$ equal parts. Use $AP = \frac{2\lambda}{5}$ and $PB = \frac{3\lambda}{5}$ .
Trusting the option label in the solution without checking the computed value. The working gives $OP = \frac{\sqrt{19}}{5}\lambda$ , so the matching option must be identified from the options list. Always match the derived expression with the option text.
Applying the cosine rule with the wrong angle or wrong side placement. In triangle $OAP$ , the included angle used is $60^\circ$ as stated in the solution. Write the cosine rule carefully before substitution.

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