WBJEE 2020 Math Solved Paper

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