AP EAPCET 06-Jul-2022 Shift 2 Paper

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