Paragraph: The line $y=x$ partitions the circle $(x-a)^2+y^2=a^2$ in two segments.
Question: What is the area of minor segment?

• Correct answer
  A

  $\frac{(\pi-2) a^2}{4}$
• B

  $\frac{(\pi-1) a^2}{4}$
• C

  $\frac{(\pi-2) a^2}{2}$
• D

  $\frac{(\pi-1) a^2}{2}$

Solution

$(x-a)^2+y^2=a^2$ Centre $:(a, 0)$, Radius $=a$
Clearly sector $O C A O$ is one quarter of circle $\therefore \quad$ Area $(\triangle O C A)=\frac{1}{2} \times a \times a=\frac{a^2}{2}$ Area of minor segment $=\frac{1}{4}($ Area of circle $)-\operatorname{Arca}(\triangle O C A)$ $=\frac{1}{4} \cdot \pi a^2-\frac{1}{2} a^2=\frac{(\pi-2) a^2}{4}$

Hence, the correct answer is option: 'A'.