TS EAMCET · Maths · Area Under Curves
The area (in square units) of the region enclosed by the two circles \(x^2+y^2=1\) and \((x-1)^2+y^2=1\) is
- A \(\frac{2 \pi}{3}+\frac{\sqrt{3}}{2}\)
- B \(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\)
- C \(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\)
- D \(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\)
Step-by-step Solution
Detailed explanation
Intersection point of two circles \(x^2+y^2=1\) \(\ldots\) (i) \((x-1)^2+y^2=1\) \(\ldots\) (ii) is given by \((x-1)^2+\left(1-x^2\right)=1\) \(\Rightarrow \quad x^2+1-2 x-x^2=0\) \(\Rightarrow \quad x=\frac{1}{2}\) From Eq. (i), \(\frac{1}{4}+y^2=1\)…
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