JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 12)
A line segment AB of length $$\lambda$$ moves such that the points A and B remain on the periphery of a circle of radius $$\lambda$$. Then the locus of the point, that divides the line segment AB in the ratio 2 : 3, is a circle of radius :
$${2 \over 3}\lambda $$
$${3 \over 5}\lambda $$
$${{\sqrt {19} } \over 7}\lambda $$
$${{\sqrt {19} } \over 5}\lambda $$
Explanation
Given, length of $A B=\lambda$
So, $A C=\frac{\lambda}{2}$ and $A M=\frac{2 \lambda}{5}$
$$ C M=A C-A M=\frac{\lambda}{2}-\frac{2 \lambda}{5}=\frac{\lambda}{10} $$
_10th_April_Morning_Shift_en_12_1.png)
Now, from $\triangle O C M$, we get
$$ \begin{array}{ll} &O M^2=C M^2+O C^2 \\\\ &\Rightarrow x^2+y^2=\left(\frac{\lambda}{10}\right)^2+\left(\lambda^2-\frac{\lambda^2}{4}\right) \\\\ &\Rightarrow x^2+y^2=\frac{\lambda^2}{100}+\frac{3 \lambda^2}{4}=\frac{\lambda^2+75 \lambda^2}{100} \\\\ &\Rightarrow x^2+y^2=\frac{76 \lambda^2}{100}=\frac{19 \lambda^2}{25} \\\\ &\therefore x^2+y^2=\left(\frac{\sqrt{19} \lambda}{5}\right)^2 \end{array} $$
So, radius is $\frac{\sqrt{19} \lambda}{5}$.
So, $A C=\frac{\lambda}{2}$ and $A M=\frac{2 \lambda}{5}$
$$ C M=A C-A M=\frac{\lambda}{2}-\frac{2 \lambda}{5}=\frac{\lambda}{10} $$
Now, from $\triangle O C M$, we get
$$ \begin{array}{ll} &O M^2=C M^2+O C^2 \\\\ &\Rightarrow x^2+y^2=\left(\frac{\lambda}{10}\right)^2+\left(\lambda^2-\frac{\lambda^2}{4}\right) \\\\ &\Rightarrow x^2+y^2=\frac{\lambda^2}{100}+\frac{3 \lambda^2}{4}=\frac{\lambda^2+75 \lambda^2}{100} \\\\ &\Rightarrow x^2+y^2=\frac{76 \lambda^2}{100}=\frac{19 \lambda^2}{25} \\\\ &\therefore x^2+y^2=\left(\frac{\sqrt{19} \lambda}{5}\right)^2 \end{array} $$
So, radius is $\frac{\sqrt{19} \lambda}{5}$.
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