Question
Question: Consider the two circles C<sub>1</sub> : x<sup>2</sup> + y<sup>2</sup> = r<sub>1</sub><sup>2</sup> a...
Consider the two circles C1 : x2 + y2 = r12 and C2 : x2 + y2 = r22 (r2< r1). Let A be a fixed point on the circle C1, set A(r1, 0) and ‘B’ be a variable point on the circle C2. Then line BA meets the circles C2 again at C. Then the set of values of
OB2 + OA2 + BC2 is –
[5r22 – 3r12, 5r22 + r12]
[3r22 – 5r12, 5r22 + r12]
[5r22 – 3r12, r22 + 5r12]
None of these
5r<sub>2</sub><sup>2</sup>–3r<sub>1</sub><sup>2</sup>,5r<sub>2</sub><sup>2</sup>+r<sub>1</sub><sup>2</sup>
Solution
Let the equation of line AB be
cosθx−r1=sinθy−0= r
The coordinates of any point on this line are
(r1 + r cos q, r sin q).
If it lies on x2 + y2 = r22. Then, we have
(r1 + r cos q)2 + (r sin q)2 = r22
r2 + 2rr1 cos q + r12 – r22 = 0 … (1)
Let AB = rB and AC = rC. Then, rB and rC are the roots of equation (1). Therefore,
rB + rC = –2 r1 cos q and rBrC = r12 – r22
\ BC2 = (rC – rB)2
= (rC + rB)2 – 4rBrC
= 4r12 cos2 q – 4r12 + 4r22
Now, OA2 + OB2 + BC2
= r12 + r22 + 4r12 cos2 q – 4r12 + 4r22
= 5r22 – 3r12 + 4r12 cos2 q
Now, 0 ฃ cos2 q ฃ 1
0 ฃ 4r12 cos2 q ฃ 4r12
5r22 – 3r12 ฃ 5r22 – 3r12 + 4r12 cos2 q
ฃ 5r22 – 3r12 + 4r12
5r22 – 3r12 ฃ OA2 + OB2 + BC2 ฃ 5r22 + r12
[using equation (2)]
OA2 + OB2 + BC2 ฮ [5r22 – 3r12 , 5r22 + r12].