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=
a−4h
2
→(3) Substitute (3) in (1) to get:12h2−4k2−24ah+9a2=0 ∴ locus : 12x2−4y2−24ax+9a2=0 i.e. a hyperbola