Let, centre =(h,k) and radius =r for the variable circle So,
using C1C2=r1+r2
for both cases we have: h2+k2=(r+a)2→(1) and (h−2a)2+k2=(r+2a)2→(2)
Eq. (2) - Eq. (1), gives : r=2a−4h→ (3)
Substitute (3) in (1) to get: 12h2−4k2−24ah+9a2=0 ∴ locus : 12x2−4y2−24ax+9a2=0 i.e. a hyperbola