The area of the circle (x - 2)^2 + (y - 3)^2 | vedclass

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(D) The equation of the circle is $(x - 2)^2 + (y - 3)^2 = 32$,which is $(x - 2)^2 + (y - 3)^2 = (4\sqrt{2})^2$. Here,the center of the circle is $(2,

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$16 \pi$

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(D) The equation of the circle is $(x - 2)^2 + (y - 3)^2 = 32$,which is $(x - 2)^2 + (y - 3)^2 = (4\sqrt{2})^2$. Here,the center of the circle is $(2, 3)$ and the radius $r = 4\sqrt{2}$. We check if the line $y = x + 1$ (or $x - y + 1 = 0$) passes through the center $(2, 3)$. Substituting $(2, 3)$ into the line equation: $2 - 3 + 1 = 0$. Since the center satisfies the equation of the line,the line is a diameter of the circle. $A$ diameter divides the circle into two equal areas. Therefore,the area of the circle lying below the line is half the total area of the circle. Area $= \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (4\sqrt{2})^2 = \frac{1}{2} \pi (32) = 16\pi$.

The area of the circle $(x - 2)^2 + (y - 3)^2 = 32$ that lies below the line $y = x + 1$ is:

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