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

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(C) 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

$3 < r < 7$

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(C) 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 $(x+1)^2 + (y+2)^2 = r^2$,the center $C_1 = (-1, -2)$ and radius $r_1 = r$.For the second circle $x^2 + y^2 - 4x - 4y + 4 = 0$,the center $C_2 = (2, 2)$ and radius $r_2 = \sqrt{2^2 + 2^2 - 4} = \sqrt{4+4-4} = 2$.The distance between the centers $C_1C_2 = \sqrt{(2 - (-1))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.Applying the condition $|r - 2| $1$) $|r - 2| $2$) $5  3$.Combining these inequalities,we get $3 < r < 7$.

If the circles $(x+1)^2+(y+2)^2=r^2$ and $x^2+y^2-4x-4y+4=0$ intersect at exactly two distinct points,then

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