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What is the image of a circle with equation $(x + 2)^2 + (y - 2)^2 = 1$ when it is rotated through $\frac{\pi}{6}$ about the origin?$\newline$(a) $(x + (\sqrt{3} + 1))^2 + (y - (\sqrt{3} - 1))^2 = 1$$\newline$(b) $(x - (\sqrt{3} + 1))^2 + (y + (\sqrt{3} - 1))^2 = 1$$\newline$(c) $(x + (\sqrt{3} - 1))^2 + (y - (\sqrt{3} + 1))^2 = 1$$\newline$(d) $(x - (\sqrt{3} - 1))^2 + (y + (\sqrt{3} + 1))^2 = 1$