Consider the given circles. S1:(x−1)2+(y−3)2=r2 And, S2:(x−4)2+(y+1)2=32 If the circle intersect at two distinct points, then $ |r_{1}-r_{2}||r−3|<√(4−1)2+(−1−3)2<(r+3) |r−3|<5<(r+3) So r+3>5 r>2 And, |r−3|<5 r−3∈(−5,5) r∈(−2,8) Thus, 2<r<8