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Question

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

$\frac{1}{2}<\mathrm{r}<7$
$3<\mathrm{r}<7$
$5<\mathrm{r}<9$
$0<\mathrm{r}<7$

Solution

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