In the xy-plane,the segment with endpoints (3 | vedclass

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(B) The center of the circle is the midpoint of the diameter: $(\frac{3-5}{2}, \frac{8+2}{2}) = (-1, 5)$.The radius squared $r^2$ is the distance from

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exactly one integral $k$

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(B) The center of the circle is the midpoint of the diameter: $(\frac{3-5}{2}, \frac{8+2}{2}) = (-1, 5)$.The radius squared $r^2$ is the distance from the center $(-1, 5)$ to $(3, 8)$ squared: $r^2 = (3 - (-1))^2 + (8 - 5)^2 = 4^2 + 3^2 = 16 + 9 = 25$.The equation of the circle is $(x + 1)^2 + (y - 5)^2 = 25$.Substitute the point $(k, 10)$ into the equation: $(k + 1)^2 + (10 - 5)^2 = 25$.$(k + 1)^2 + 5^2 = 25$.$(k + 1)^2 + 25 = 25$.$(k + 1)^2 = 0$.$k + 1 = 0$,so $k = -1$.Since $k = -1$ is an integer,there is exactly one integral value of $k$.

In the $xy$-plane,the segment with endpoints $(3, 8)$ and $(-5, 2)$ is the diameter of a circle. The point $(k, 10)$ lies on the circle for:

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