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 :
- A \((0, \pm 1)\)
- B \(\left(0, \pm \frac{\sqrt{3}}{2}\right)\)
- C \(\left(\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)\)
- D \(( \pm 1,0)\)
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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