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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 –

A

[5r22 – 3r12, 5r22 + r12]

B

[3r22 – 5r12, 5r22 + r12]

C

[5r22 – 3r12, r22 + 5r12]

D

None of these

Answer

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>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>

Explanation

Solution

Let the equation of line AB be

xr1cosθ=y0sinθ\frac{x - r_{1}}{\cos\theta} = \frac{y - 0}{\sin\theta}= 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].