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(D) For the first circle $x^{2}+y^{2}-10x-10y+41=0$:Center $C_{1} = (5, 5)$,Radius $r_{1} = \sqrt{5^{2}+5^{2}-41} = \sqrt{25+25-41} = \sqrt{9} = 3$.Fo

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(D) For the first circle $x^{2}+y^{2}-10x-10y+41=0$:Center $C_{1} = (5, 5)$,Radius $r_{1} = \sqrt{5^{2}+5^{2}-41} = \sqrt{25+25-41} = \sqrt{9} = 3$.For the second circle $x^{2}+y^{2}-24x-10y+160=0$:Center $C_{2} = (12, 5)$,Radius $r_{2} = \sqrt{12^{2}+5^{2}-160} = \sqrt{144+25-160} = \sqrt{9} = 3$.The distance between the centers $C_{1}$ and $C_{2}$ is $d = \sqrt{(12-5)^{2}+(5-5)^{2}} = \sqrt{7^{2}+0^{2}} = 7$.The minimum distance between the two circles is given by $d - (r_{1} + r_{2}) = 7 - (3 + 3) = 7 - 6 = 1$.

The minimum distance between any two points $P_{1}$ and $P_{2}$,where $P_{1}$ lies on the first circle and $P_{2}$ lies on the second circle,for the given equations:
$x^{2}+y^{2}-10x-10y+41=0$
$x^{2}+y^{2}-24x-10y+160=0$
is .........

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The minimum distance between any two points $P_{1}$ and $P_{2}$,where $P_{1}$ lies on the first circle and $P_{2}$ lies on the second circle,for the given equations:$x^{2}+y^{2}-10x-10y+41=0$$x^{2}+y^{2}-24x-10y+160=0$is .........