The centre of the circle x2+y2+2αx+c=0 is (−α,0). As the circle x2+y2+2αx+c=0 lies inside the circle x2+y2+2βx+c=0, then ‌(−α)2+02+2β(−α)+c<0 ‌α2−2αβ+c<0 ‌α2−2αβ+β2<β2−c .....(i) (α−β)2<β2−c Also, radius of circle x2+y2+2αx+c=0 less than radius of the circle x2+y2+2βx+c=0 ‌√α2−c<√β2−c ⇒α2−c<β2−c⇒α<β ‌α2−2αβ<0....(ii) From Eqs. (i) and (ii), ‌c>0 ‌⇒‌‌−2αβ<0 ‌αβ>0