The least distance of the point A(10, 7) from | vedclass

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(A) The equation of the circle is $x^2 + y^2 - 4x - 2y - 20 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -1$,and $c = -20

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$5, 15$

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(A) The equation of the circle is $x^2 + y^2 - 4x - 2y - 20 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -1$,and $c = -20$. The center $C$ is $(-g, -f) = (2, 1)$. The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-1)^2 - (-20)} = \sqrt{4 + 1 + 20} = \sqrt{25} = 5$. The distance $AC = \sqrt{(10 - 2)^2 + (7 - 1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$. The least distance $AM = AC - r = 10 - 5 = 5$. Since $MM'$ is the diameter passing through $C$,$M'$ is the point on the circle farthest from $A$. The maximum distance $AM' = AC + r = 10 + 5 = 15$. Thus,the lengths are $5$ and $15$ units.

The least distance of the point $A(10, 7)$ from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$ is the length of segment $AM$. If $MM'$ is the diameter of the circle,then the lengths of $AM$ and $AM'$ are respectively . . . . . . , . . . . . . units.

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