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

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(B) For the first circle $(x - 1)^2 + (y - 3)^2 = r^2$,the center $C_1 = (1, 3)$ and radius $R_1 = r$.For the second circle $x^2 + y^2 - 8x + 2y + 8 =

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$2 < r < 8$

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(B) For the first circle $(x - 1)^2 + (y - 3)^2 = r^2$,the center $C_1 = (1, 3)$ and radius $R_1 = r$.For the second circle $x^2 + y^2 - 8x + 2y + 8 = 0$,the center $C_2 = (-(-8)/2, -(2)/2) = (4, -1)$ and radius $R_2 = \sqrt{g^2 + f^2 - c} = \sqrt{(-4)^2 + (1)^2 - 8} = \sqrt{16 + 1 - 8} = \sqrt{9} = 3$.The distance between the centers $d = \sqrt{(4 - 1)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.Two circles intersect at two distinct points if $|R_1 - R_2| Substituting the values: $|r - 3| From $r + 3 > 5$,we get $r > 2$.From $|r - 3| Combining $r > 2$ and $r < 8$,we get $2 < r < 8$.

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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