Consider the two circles C 1 : x 2 + y 2 = r 1 2 and C 2 : x 2 + y 2 = r 2 2 (r 2 < r 1 ). Let A be a fixed point on the circle C 1 , set A(r 1 , 0) and ‘B’ be a variable point on the circle C 2 . Then line BA meets the circles C 2 again at C. Then the set of values of OB 2 + OA 2 + BC 2 is -
Text Solution
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Let the equation of line AB be
= r
The coordinates of any point on this line are
(r 1 + r cos θ , r sin θ ).
If it lies on x 2 + y 2 = r 2 2 . Then, we have
(r 1 + r cos θ ) 2 + (r sin θ ) 2 = r 2 2
⇒ r 2 + 2rr 1 cos θ + r 1 2 – r 2 2 = 0 … (1)
Let AB = r B and A C = r C . Then, r B and r C are the roots of equation (1). Therefore,
r B + r C = –2 r 1 cos θ and r B r C = r 1 2 – r 2 2
∴ BC 2 = (r C – r B ) 2
= (r C + r B ) 2 – 4r B r C
= 4r 1 2 cos 2 θ – 4r 1 2 + 4r 2 2
Now, OA 2 + OB 2 + BC 2
= r 1 2 + r 2 2 + 4r 1 2 cos 2 θ – 4r 1 2 + 4r 2 2
= 5r 2 2 – 3r 1 2 + 4r 1 2 cos 2 θ
Now, 0 ≤ cos 2 θ ≤ 1
⇒ 0 ≤ 4r 1 2 cos 2 θ ≤ 4r 1 2
⇒ 5r 2 2 – 3r 1 2 ≤ 5r 2 2 – 3r 1 2 + 4r 1 2 cos 2 θ
≤ 5r 2 2 – 3r 1 2 + 4r 1 2
⇒ 5r 2 2 – 3r 1 2 ≤ OA 2 + OB 2 + BC 2 ≤ 5r 2 2 + r 1 2
[using equation (2)]
⇒ OA 2 + OB 2 + BC 2 ∈ [5r 2 2 – 3r 1 2 , 5r 2 2 + r 1 2 ].
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