The radius of the circle (x - 1)(x - 3) + (y | vedclass

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(B) The given equation of the circle is $(x - 1)(x - 3) + (y - 2)(y - 4) = 0$.This is in the form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$,which

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$\sqrt{2}$

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(B) The given equation of the circle is $(x - 1)(x - 3) + (y - 2)(y - 4) = 0$.This is in the form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$,which represents a circle with the diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$.Here,the endpoints of the diameter are $(1, 2)$ and $(3, 4)$.The length of the diameter $d = \sqrt{(3 - 1)^2 + (4 - 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.The radius $r = \frac{d}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$.

The radius of the circle $(x - 1)(x - 3) + (y - 2)(y - 4) = 0$ is

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