Let a circle C_1 equiv x^2 + y^2 - 4x + 6y + | vedclass

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(D) For circle $C_1: x^2 + y^2 - 4x + 6y + 1 = 0$,the centre is $(2, -3)$ and radius $r = \sqrt{2^2 + (-3)^2 - 1} = \sqrt{4 + 9 - 1} = \sqrt{12} = 2\s

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$8\pi + 6\sqrt{3}$

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(D) For circle $C_1: x^2 + y^2 - 4x + 6y + 1 = 0$,the centre is $(2, -3)$ and radius $r = \sqrt{2^2 + (-3)^2 - 1} = \sqrt{4 + 9 - 1} = \sqrt{12} = 2\sqrt{3}$.Since the centre of $C_2$ is the reflection of $(2, -3)$ about the $x$-axis,the centre of $C_2$ is $(2, 3)$. The radius of $C_2$ is also $2\sqrt{3}$.The distance between the centres $d = \sqrt{(2-2)^2 + (3 - (-3))^2} = \sqrt{0^2 + 6^2} = 6$.The area of $C_1$ not common with $C_2$ is the total area of $C_1$ minus the area of the intersection of $C_1$ and $C_2$.The area of intersection of two circles with equal radii $r$ and distance $d$ between centres is $2r^2 \cos^{-1}(\frac{d}{2r}) - \frac{d}{2} \sqrt{4r^2 - d^2}$.Here $r = 2\sqrt{3}$,$r^2 = 12$,$d = 6$. $\frac{d}{2r} = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}$.Area of intersection $= 2(12) \cos^{-1}(\frac{\sqrt{3}}{2}) - \frac{6}{2} \sqrt{4(12) - 36} = 24(\frac{\pi}{6}) - 3\sqrt{48-36} = 4\pi - 3\sqrt{12} = 4\pi - 6\sqrt{3}$.Area of $C_1$ not common with $C_2 = 	ext{Area}(C_1) - 	ext{Area of intersection} = 12\pi - (4\pi - 6\sqrt{3}) = 8\pi + 6\sqrt{3}$.

Let a circle $C_1 \equiv x^2 + y^2 - 4x + 6y + 1 = 0$ and circle $C_2$ be such that its centre is the image of the centre of $C_1$ about the $x$-axis and the radius of $C_2$ is equal to the radius of $C_1$. Then,the area of $C_1$ which is not common with $C_2$ is:

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