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
Step-by-Step Solution
Key Concept: Circle 2: centre $(2,2)$, radius $2$. Circle 1: centre $(-1,-2)$, radius $r$. Distance $CC'=\sqrt{9+16}=5$. For two intersection points: $|r-2|<5<r+2$, giving $r>3$ and $r<7$.
$CC'=5$, $r_2=2$. Conditions: $-3<r<7$ and $r>3\Rightarrow3<r<7$.
Correct Answer: 3