Question

Find the area of the circle \((x-2)^2+y^2=4\).

Hint:

This is a circle of radius 2; area = π×2² (or verify via ∫√(4−X²)dX).

Answer 1

Step 1: Identifying the circle's parameters:
\((x-2)^2+y^2=4=2^2\) is a circle centred at \((2,0)\) with radius \(r=2\).

Step 2: Setting up the area via integration:
Shifting coordinates \(X=x-2\) doesn't change area, so consider \(X^2+y^2=4\Rightarrow y=\pm\sqrt{4-X^2}\); area \(=2\displaystyle\int_{-2}^{2}\sqrt{4-X^2}\,dX\) (the factor 2 for upper and lower half).

Step 3: Using the standard integral formula:
\(\displaystyle\int\sqrt{a^2-X^2}\,dX=\dfrac{X}{2}\sqrt{a^2-X^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{X}{a}+C\), with \(a=2\).

Step 4: Evaluating from −2 to 2:
\(\left[\dfrac{X}{2}\sqrt{4-X^2}+2\sin^{-1}\dfrac X2\right]_{-2}^{2}=\left(0+2\sin^{-1}1\right)-\left(0+2\sin^{-1}(-1)\right)=2\cdot\dfrac\pi2-2\cdot\left(-\dfrac\pi2\right)=\pi+\pi=2\pi\).

Step 5: Applying the factor of 2:
Area \(=2\times2\pi=4\pi\).

Final Answer:
\[ \boxed{\text{Area}=4\pi \text{ sq. units}} \]