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

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(D) The given equation of the circle is $x^2 + y^2 - 4x - 2y - 20 = 0$. Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $

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$5$

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(D) The given equation of the circle is $x^2 + y^2 - 4x - 2y - 20 = 0$. Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -1$,and $c = -20$. The center of the circle is $C = (-g, -f) = (2, 1)$. The radius $r$ is given by $\sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-1)^2 - (-20)} = \sqrt{4 + 1 + 20} = \sqrt{25} = 5$. Let the point be $P = (10, 7)$. The distance $d$ between the point $P$ and the center $C$ is $\sqrt{(10 - 2)^2 + (7 - 1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$. Since the distance $d = 10$ is greater than the radius $r = 5$,the point lies outside the circle. The least distance of the point from the circle is $d - r = 10 - 5 = 5$ units.

The least distance of the point $(10, 7)$ from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$ is

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