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QJEEM23Ma72eb: JEE Main 2023 Mathematics - Class 11 JEEM23Ma72eb Solution: A line segment AB of length moves such that the points A and B…

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

Multiple Choice Options

  • A 23λ{2 \over 3}\lambda
  • B 35λ{3 \over 5}\lambda
  • C 197λ{{\sqrt {19} } \over 7}\lambda
  • D 195λ{{\sqrt {19} } \over 5}\lambda

Step-by-Step Verified Solution

Given, length of AB=λA B=\lambda

So, AC=λ2A C=\frac{\lambda}{2} and AM=2λ5A M=\frac{2 \lambda}{5}

CM=AC−AM=λ2−2λ5=λ10C M=A C-A M=\frac{\lambda}{2}-\frac{2 \lambda}{5}=\frac{\lambda}{10}

JEE Main 2023 (Online) 10th April Morning Shift Mathematics - Circle Question 32 English Explanation

Now, from △OCM\triangle O C M , we get

amp;OM2=CM2+OC2amp;⇒x2+y2=(λ10)2+(λ2−λ24)amp;⇒x2+y2=λ2100+3λ24=λ2+75λ2100amp;⇒x2+y2=76λ2100=19λ225amp;∴x2+y2=(19λ5)2\begin{array}{ll}&O M^2=C M^2+O C^2 \\\\&\Rightarrow x^2+y^2=\left(\frac{\lambda}{10}\right)^2+\left(\lambda^2-\frac{\lambda^2}{4}\right) \\\\&\Rightarrow x^2+y^2=\frac{\lambda^2}{100}+\frac{3 \lambda^2}{4}=\frac{\lambda^2+75 \lambda^2}{100} \\\\&\Rightarrow x^2+y^2=\frac{76 \lambda^2}{100}=\frac{19 \lambda^2}{25} \\\\&\therefore x^2+y^2=\left(\frac{\sqrt{19} \lambda}{5}\right)^2\end{array}

So, radius is 19λ5\frac{\sqrt{19} \lambda}{5} .

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