Condition 1: The centre of the two circles are (1,1) and (9,1) . The circles are on opposite sides of the line 3x+4y−λ=0 Put x=1,y=1 in the equation of line, 3(1)+4(1)−λ=0⇒7−λ=0 Now, put x=9,y=1 in the equation of line, 3(9)+4(1)−λ=0 Then, (7−λ)(27+4−λ)<0 ⇒(λ−7)(λ−31)<0 λ∈(7,31)...(i) Condition 2: Perpendicular distance from centre on line ≥ radius of circle. For x2+y2−2x−2y=1, ⇒
|3+4−λ|
5
≥1 ⇒|λ−7|≥5 ⇒λ≥12 or λ⇒2 . . . (ii) For x2+y2−18x−2y+78=0
|27+4−λ|
5
≥2 ⇒λ≥41 or λ≤21 . . . (iii) Intersection of (1),(2) and (3) gives λ∈[12,21]