JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 13)
Let the locus of the midpoints of the chords of the circle $x^2+(y-1)^2=1$ drawn from the origin intersect the line $x+y=1$ at $\mathrm{P}$ and $\mathrm{Q}$. Then, the length of $\mathrm{PQ}$ is :
$\frac{1}{2}$
1
$\frac{1}{\sqrt{2}}$
$\sqrt{2}$
Explanation
Let mid-point is $(x, y)$
$$ \begin{aligned} & x^2+y^2-2 y=0 \\\\ & x x_1+y y_1-\left(y+y_1\right)=x_1^2+y_1^2-2 y_1 \end{aligned} $$
It is passing through origin
$$ \begin{aligned} & \text { So, } 0+0-\left(0+y_1\right)=x_1^2+y_1^2-2 y_1 \\\\ & \Rightarrow -y_1=x_1^2+y_1^2-2 y_1 \\\\ & \Rightarrow x_1^2+y_1^2-y_1=0 \end{aligned} $$
$$ x^2+y^2-y=0 $$ ............ (1)
$\because$ It intersects the line $x+y=1$
So put $x=1-y$ in equation (1)
$$ \begin{aligned} & (1-y)^2+y^2-y=0 \\\\ & 2 y^2-3 y+1=0 \end{aligned} $$
$\begin{aligned} & \Rightarrow (y-1)(2 y-1)=0 \\\\ & \Rightarrow y=1, \frac{1}{2} \\\\ & \therefore P(0,1) \text { and } Q\left(\frac{1}{2}, \frac{1}{2}\right)\end{aligned}$
So, $P Q=\sqrt{\left(\frac{1}{2}-0\right)^2+\left(\frac{1}{2}-1\right)^2}=\frac{1}{\sqrt{2}}$
$$ \begin{aligned} & x^2+y^2-2 y=0 \\\\ & x x_1+y y_1-\left(y+y_1\right)=x_1^2+y_1^2-2 y_1 \end{aligned} $$
It is passing through origin
$$ \begin{aligned} & \text { So, } 0+0-\left(0+y_1\right)=x_1^2+y_1^2-2 y_1 \\\\ & \Rightarrow -y_1=x_1^2+y_1^2-2 y_1 \\\\ & \Rightarrow x_1^2+y_1^2-y_1=0 \end{aligned} $$
$$ x^2+y^2-y=0 $$ ............ (1)
$\because$ It intersects the line $x+y=1$
So put $x=1-y$ in equation (1)
$$ \begin{aligned} & (1-y)^2+y^2-y=0 \\\\ & 2 y^2-3 y+1=0 \end{aligned} $$
$\begin{aligned} & \Rightarrow (y-1)(2 y-1)=0 \\\\ & \Rightarrow y=1, \frac{1}{2} \\\\ & \therefore P(0,1) \text { and } Q\left(\frac{1}{2}, \frac{1}{2}\right)\end{aligned}$
So, $P Q=\sqrt{\left(\frac{1}{2}-0\right)^2+\left(\frac{1}{2}-1\right)^2}=\frac{1}{\sqrt{2}}$
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