Published by:
CGP EDU Academic Team
Published on: August 13, 2026
Consider two circles C 1 : x 2 + y 2 – 1 = 0 and C 2 : x 2 + y 2 – 2 = 0. Let A(1,0) be a fixed point on the circle C 1 and B be any variable point on the circle C 2 . The line BA meets the curve C 2 again at C.
Which of the following alternative(s) is/are correct ?
A
OA 2 + OB 2 + BC 2 ∈ [7, 11], where O is the origin. (
B
OA 2 + OB 2 + BC 2 ∈ [4, 7], where O is the origin.
C
Locus of midpoint of AB is a circle of radius 
.
.
D
Locus of midpoint of AB is a circle of area 
.
.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(a,c,d)We have maximum BC = 
and minimum BC = 2
∴ OA 2 + OB 2 + BC 2 ∈ [7, 11]
Let M be the midpoint of AB.
⇒ M ≡ 
= (h, k)
∴ sin θ = 
, cos θ = 
∴ Now on squaring & adding, we get
⇒ 4k 2 + (2h – 1) 2 = 2
∴ Locus of M (h, k) is 
+ y 2 = 
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