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Circle question

2023 · 24 Jan · Shift 2 · Q29
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  5. /2023 · 24 Jan · Shift 2 · Q29

Circle question

2023 · 24 Jan · Shift 2 · Q29

JEE MainMathematicsCircleMCQ+4 / −1
The locus of the mid points of the chords of the circle C1:(x−4)2+(y−5)2=4{C_1}:{(x - 4)^2} + {(y - 5)^2} = 4C1​:(x−4)2+(y−5)2=4 which subtend an angle θi{\theta _i}θi​ at the centre of the circle C1C_1C1​, is a circle of radius rir_iri​. If θ1=π3,θ3=2π3{\theta _1} = {\pi \over 3},{\theta _3} = {{2\pi } \over 3}θ1​=3π​,θ3​=32π​ and r12=r22+r32r_1^2 = r_2^2 + r_3^2r12​=r22​+r32​, then θ2{\theta _2}θ2​ is equal to :
  1. A
    π2{\pi \over 2}2π​
  2. B
    π4{\pi \over 4}4π​
  3. C
    3π4{{3\pi } \over 4}43π​
  4. D
    π6{\pi \over 6}6π​
View written solutionFree

Correct answer: A

  1. Given circle and its radius

The circle is

(x−4)2+(y−5)2=4(x-4)^2+(y-5)^2=4(x−4)2+(y−5)2=4

So its centre is (4,5)(4,5)(4,5) and radius is

R=2.R=2.R=2.
  1. Locus of midpoints of chords subtending a fixed angle at the centre

For a chord of a circle of radius RRR subtending angle θ\thetaθ at the centre, if ddd is the perpendicular distance of the chord from the centre, then

d=Rcos⁡θ2.d=R\cos\frac{\theta}{2}.d=Rcos2θ​.

Since the midpoint of a chord lies on the perpendicular from the centre to the chord, all such midpoints are at a fixed distance ddd from the centre. Hence their locus is a circle concentric with the given circle, of radius

r=Rcos⁡θ2.r=R\cos\frac{\theta}{2}.r=Rcos2θ​.

So here,

ri=2cos⁡θi2.r_i=2\cos\frac{\theta_i}{2}.ri​=2cos2θi​​.

Therefore,

ri2=4cos⁡2θi2.r_i^2=4\cos^2\frac{\theta_i}{2}.ri2​=4cos22θi​​.
  1. Compute r12r_1^2r12​ and r32r_3^2r32​

Given

θ1=π3,θ3=2π3.\theta_1=\frac{\pi}{3},\qquad \theta_3=\frac{2\pi}{3}.θ1​=3π​,θ3​=32π​.

So,

r12=4cos⁡2π6=4(32)2=4⋅34=3.r_1^2=4\cos^2\frac{\pi}{6}=4\left(\frac{\sqrt3}{2}\right)^2=4\cdot\frac34=3.r12​=4cos26π​=4(23​​)2=4⋅43​=3.

And,

r32=4cos⁡2π3=4(12)2=1.r_3^2=4\cos^2\frac{\pi}{3}=4\left(\frac12\right)^2=1.r32​=4cos23π​=4(21​)2=1.
  1. Use the relation r12=r22+r32r_1^2=r_2^2+r_3^2r12​=r22​+r32​

Given,

r12=r22+r32.r_1^2=r_2^2+r_3^2.r12​=r22​+r32​.

Substitute the values:

3=r22+13=r_2^2+13=r22​+1

Hence,

r22=2.r_2^2=2.r22​=2.

Now,

r22=4cos⁡2θ22=2r_2^2=4\cos^2\frac{\theta_2}{2}=2r22​=4cos22θ2​​=2

So,

cos⁡2θ22=12.\cos^2\frac{\theta_2}{2}=\frac12.cos22θ2​​=21​.

Thus,

θ22=π4\frac{\theta_2}{2}=\frac{\pi}{4}2θ2​​=4π​

for the principal value relevant here, giving

θ2=π2.\theta_2=\frac{\pi}{2}.θ2​=2π​.
  1. Check options
  • A: π2\frac{\pi}{2}2π​ ✅
  • B: π4\frac{\pi}{4}4π​
  • C: 3π4\frac{3\pi}{4}43π​
  • D: π6\frac{\pi}{6}6π​

Therefore, the correct option is A.

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