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

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(D) For two circles to intersect at two distinct points,the distance between their centers $C_1C_2$ must satisfy the condition: $|r_1 - r_2| < C_1C_2

Vedclass · Accepted Answer

$2 < r < 8$

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(D) For two circles to intersect at two distinct points,the distance between their centers $C_1C_2$ must satisfy the condition: $|r_1 - r_2| 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 $C_2 = (-g, -f) = (4, -1)$ 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 $C_1(1, 3)$ and $C_2(4, -1)$ is $C_1C_2 = \sqrt{(4-1)^2+(-1-3)^2} = \sqrt{3^2+(-4)^2} = \sqrt{9+16} = 5$. Substituting these values into the condition: $|r - 3| From $5  2$. From $|r - 3| Combining $r > 2$ and $0 < r < 8$,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 in two different points,then what can we conclude about $r$?

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