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Mathematics · Circles

JEE Main 2024 — 1 February, Shift 2 — Question 20

Let the locus of the mid points of the chords of circle x2+(y−1)2=1x^{2}+(y-1)^{2}=1 drawn from the origin intersect the line x+y=1x+y=1 at PP and QQ. Then, the length of PQP Q is :

  1. Option A:

    12\frac{1}{\sqrt{2}}

    Correct
  2. Option B:

    2\sqrt{2}

  3. Option C:

    12\frac{1}{2}

  4. Option D:

    1

Answer: A

Step-by-step solution

figure

mOM⋅mCM=−1\mathrm{m}_{\mathrm{OM}} \cdot \mathrm{m}_{\mathrm{CM}}=-1

kh⋅k−1 h=−1\frac{\mathrm{k}}{\mathrm{h}} \cdot \frac{\mathrm{k}-1}{\mathrm{~h}}=-1

∴\therefore locus is x2+y(y−1)=0\mathrm{x}^{2}+\mathrm{y}(\mathrm{y}-1)=0

x2+y2−y=0x^{2}+y^{2}-y=0

figure

p=∣1/22∣p=\left|\frac{1 / 2}{\sqrt{2}}\right| \quad

p=122 \mathrm{p}=\frac{1}{2 \sqrt{2}}

PQ=2r2−p2\mathrm{PQ}=2 \sqrt{\mathrm{r}^{2}-\mathrm{p}^{2}}

=214−18=12=2 \sqrt{\frac{1}{4}-\frac{1}{8}}=\frac{1}{\sqrt{2}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 2
Subject
Mathematics
Chapter
Circles
Topic
Chords connected with a Circle
Let the locus of the mid points of the chords of circle x 2 +(y-1) 2… | JEE Main 2024 PYQ with Solution · DhiX AI