Verified Solution • 2025–26 Session moderate

QJEEM24Mafad0: JEE Main 2024 Mathematics - Class 11 JEEM24Mafad0 Solution: Let the locus of the midpoints of the chords of the circle…

Let the locus of the midpoints of the chords of the circle x2+(y−1)2=1x^2+(y-1)^2=1 drawn from the origin intersect the line x+y=1x+y=1 at P\mathrm{P} and Q\mathrm{Q} . Then, the length of PQ\mathrm{PQ} is :

Multiple Choice Options

  • A 12\frac{1}{2}
  • B 1
  • C 12\frac{1}{\sqrt{2}}
  • D 2\sqrt{2}

Step-by-Step Verified Solution

Let mid-point is (x,y)(x, y)

amp;x2+y2−2y=0amp;xx1+yy1−(y+y1)=x12+y12−2y1\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

amp; So, 0+0−(0+y1)=x12+y12−2y1amp;⇒−y1=x12+y12−2y1amp;⇒x12+y12−y1=0\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}

x2+y2−y=0x^2+y^2-y=0 ............ (1)

∵\because It intersects the line x+y=1x+y=1

So put x=1−yx=1-y in equation (1)

amp;(1−y)2+y2−y=0amp;2y2−3y+1=0\begin{aligned}& (1-y)^2+y^2-y=0 \\\\& 2 y^2-3 y+1=0\end{aligned}

amp;⇒(y−1)(2y−1)=0amp;⇒y=1,12amp;∴P(0,1) and Q(12,12)\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, PQ=(12−0)2+(12−1)2=12P Q=\sqrt{\left(\frac{1}{2}-0\right)^2+\left(\frac{1}{2}-1\right)^2}=\frac{1}{\sqrt{2}}

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