(2,π3)(2, \dfrac{\pi}{3})(2,3π) x=rcosθ=2cosπ3=2⋅12=1x = r\cos\theta = 2\cos\dfrac{\pi}{3} = 2 \cdot \dfrac{1}{2} = 1x=rcosθ=2cos3π=2⋅21=1 y=rsinθ=2sinπ3=2⋅32=3y = r\sin\theta = 2\sin\dfrac{\pi}{3} = 2 \cdot \dfrac{\sqrt{3}}{2} = \sqrt{3}y=rsinθ=2sin3π=2⋅23=3 (1,3)(1, \sqrt{3})(1,3)を中心とし、半径２の円なので ∴(x−1)2+(y−3)2=4\therefore (x-1)^2 + (y-\sqrt{3})^2 = 4∴(x−1)2+(y−3)2=4