The sum of the minimum and maximum distances | vedclass

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(A) The equation of the circle is $x^2+y^2+4x-10y-7=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=2$,$f=-5$,and $c=-7$. The center of the ci

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$20$

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(A) The equation of the circle is $x^2+y^2+4x-10y-7=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=2$,$f=-5$,and $c=-7$. The center of the circle is $C(-g, -f) = (-2, 5)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{2^2+(-5)^2-(-7)} = \sqrt{4+25+7} = \sqrt{36} = 6$. Let $P$ be the point $(4, -3)$. The distance $d$ between the center $C(-2, 5)$ and point $P(4, -3)$ is $d = \sqrt{(4 - (-2))^2 + (-3 - 5)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. The minimum distance from the point to the circle is $|d - r| = |10 - 6| = 4$. The maximum distance from the point to the circle is $d + r = 10 + 6 = 16$. The sum of the minimum and maximum distances is $4 + 16 = 20$.

The sum of the minimum and maximum distances of the point $(4,-3)$ to the circle $x^2+y^2+4x-10y-7=0$ is

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