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CUET · MATHS · PYQ PAPER 2023

Consider the curves, \(x^2+y^2=1\) and \((x-1)^2+y^2=1\).
Points of intersection of two curves are :

  1. A \((0, \pm 1)\)
  2. B \(\left(0, \pm \frac{\sqrt{3}}{2}\right)\)
  3. C \(\left(\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)\)
  4. D \(( \pm 1,0)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left(\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)\)

Step-by-step Solution

Detailed explanation

\(x^2+y^2=1\) \((x-1)^2+y^2=1\) \(x^2-(x-1)^2=0\) \(x^2-(x^2-2x+1)=0\) \(2x-1=0 \implies x=\frac{1}{2}\) \((\frac{1}{2})^2+y^2=1\) \(\frac{1}{4}+y^2=1\) \(y^2=\frac{3}{4} \implies y=\pm\frac{\sqrt{3}}{2}\) Points of intersection:…
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