If the two circles (x-1)^2+(y-3)^2=r^2 and x^ | vedclass

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(A) The equations of the given circles are:$S_1: (x-1)^2 + (y-3)^2 = r^2$ (Center $C_1 = (1, 3)$,Radius $r_1 = r$)$S_2: x^2 + y^2 - 8x + 2y + 8 = 0$Co

Vedclass · Accepted Answer

$2 < r < 8$

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(A) The equations of the given circles are:$S_1: (x-1)^2 + (y-3)^2 = r^2$ (Center $C_1 = (1, 3)$,Radius $r_1 = r$)$S_2: x^2 + y^2 - 8x + 2y + 8 = 0$Completing the square for $S_2$:$(x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1$$(x-4)^2 + (y+1)^2 = 9 = 3^2$ (Center $C_2 = (4, -1)$,Radius $r_2 = 3$)For two circles to intersect at two distinct points,the distance between their centers $d = C_1C_2$ must satisfy $|r_1 - r_2| Calculate $d = \sqrt{(4-1)^2 + (-1-3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.Thus,$|r - 3| From $r + 3 > 5$,we get $r > 2$.From $|r - 3| Combining $r > 2$ and $-2 Therefore,option $A$ is correct.

If the two circles $(x-1)^2+(y-3)^2=r^2$ and $x^2+y^2-8x+2y+8=0$ intersect at two distinct points,then

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