The equation of the circle of minimum radius which contains the three circles
x 2 + y 2 – 4y – 5 = 0 … (1)
x 2 + y 2 + 12x + 4y + 31 = 0 … (2)
and x 2 + y 2 + 6x + 12y + 36 = 0 … (3)
is
+ 
= 
+ 
= 
+ 
= 
Text Solution
Verified by ExpertsC
The coordinates of the centres and radii of three given circles are as given below :
Centre Radius
Circle (1) 
Circle (2) 
Circle (3) 
Let C (h, k) be the centre of the circle passing through the centres of the circles (1), (2) and (3). Then,
⇒ (h – 0) 2 + (k – 2) 2 = (h + 6) 2 + (k + 2) 2 = (h + 3)
2 + (k + 6) 2
⇒ –4k + 4 = 12h + 4k + 40 = 6h + 12k + 45
⇒ 12h + 8k + 36 = 0 and 6h – 8k – 5 = 0
⇒ 3h + 2k + 9 = 0 and 6h – 8k – 5 = 0
⇒ h = 
, k = 
Now, 
Thus, required circle has its centre at 
and radius = CP = 
.
Hence, its equation is
+ 
= 
.
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