The radius of the circle (x - 5)(x - 1) + (y | vedclass

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(C) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Comparing

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$5/2$

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(C) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Comparing this with the given equation $(x - 5)(x - 1) + (y - 7)(y - 4) = 0$,the endpoints of the diameter are $(5, 7)$ and $(1, 4)$.The length of the diameter $d$ is the distance between these two points:$d = \sqrt{(5 - 1)^2 + (7 - 4)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.The radius $r$ is half of the diameter:$r = \frac{d}{2} = \frac{5}{2}$.

The radius of the circle $(x - 5)(x - 1) + (y - 7)(y - 4) = 0$ is

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