It is given that the circles x2+y2−4x+6y+13−a2=0 and x2+y2−10x−2y+17=0 intersect in two distinct points, the $|r_{1}-r_{2}| where, c1=(2,−3),c2=(5,1),r1=|a| and r2=3 So, ||a|−3|<√9+16<|a|+3 ⇒||a|−3|<5<|a|+3 So, a∈(−∞,−2)∪(2,∞) and a∈(−8,8) Therefore, a∈(−8,−2)∪(2,8)