(b,d)Given circles are

x 2 + y 2 = a 2 .........(1)

and (x – 2a) 2 + y 2 = a 2 .........(2)

Let A and B be the centres and r 1 and r 2 the radii of the circles (1) and (2) respectively. Then

A ≡ (0, 0), B ≡ (2a, 0), r 1 = a, r 2 = a

Now AB = = 2a = r 1 + r 2

Hence the two circles touch each other extenally.

Let the equation of the circle having same radius ‘a’ and touching the circles (1) and (2) be

(x – α ) 2 + (y – β ) 2 = a 2 ..........(3)

Its centre C is ( α, β ) and radius r 3 = a

Since circle (3) touches the circle (1),

AC = r 1 + r 3 = 2a. [Here AC ≠ |r 1 – r 3 | as r 1 – r 3 = a – a = 0]

⇒ AC 2 = 4a 2 ⇒ α 2 + β 2 = 4a 2 ..........(4)

Again since circle (3) touches the circle (2)

BC = r 2 + r 3 ⇒ BC 2 = (r 2 )

⇒ (2a – α ) 2 + β 2 = (a + a) 2 ⇒ α 2 + β 2 – 4a α = 0 ⇒ 4a 2 – 4a α = 0 [from (4)]

⇒ α = a and from (4), we have β = ± a.

Hence, the required circles are

(x – a) 2 + (y a ) 2 = a 2

or x 2 + y 2 – 2ax 2 ay + 3a 2 = 0.