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QJEEM23Ma9700: JEE Main 2023 Mathematics - Class 11 JEEM23Ma9700 Solution: The locus of the mid points of the chords of the circle which…

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} = 4 which subtend an angle θi{\theta _i} at the centre of the circle C1C_1 , is a circle of radius rir_i . If θ1=π3,θ3=2π3{\theta _1} = {\pi \over 3},{\theta _3} = {{2\pi } \over 3} and r12=r22+r32r_1^2 = r_2^2 + r_3^2 , then θ2{\theta _2} is equal to :

Multiple Choice Options

  • A π2{\pi \over 2}
  • B π4{\pi \over 4}
  • C 3π4{{3\pi } \over 4}
  • D π6{\pi \over 6}

Step-by-Step Verified Solution

JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Circle Question 37 English Explanation

∴cos⁡(θ12)=ri2⇒ri=2cos⁡(θi2)\therefore \cos \left(\frac{\theta_{1}}{2}\right)=\frac{r_{i}}{2} \Rightarrow r_{i}=2 \cos \left(\frac{\theta_{i}}{2}\right)

Given r12=r22+r33r_{1}^{2}=r_{2}^{2}+r_{3}^{3}

⇒(cos⁡(θ12))2=(cos⁡(θ22))2+(cos⁡(θ32))2\Rightarrow\left(\cos \left(\frac{\theta_{1}}{2}\right)\right)^{2}=\left(\cos \left(\frac{\theta_{2}}{2}\right)\right)^{2}+\left(\cos \left(\frac{\theta_{3}}{2}\right)\right)^{2}

⇒34=cos⁡2(θ22)+14\Rightarrow \frac{3}{4}=\cos ^{2}\left(\frac{\theta_{2}}{2}\right)+\frac{1}{4}

⇒cos⁡2(θ22)=12\Rightarrow \cos ^{2}\left(\frac{\theta_{2}}{2}\right)=\frac{1}{2}

⇒θ22=π4\Rightarrow \frac{\theta_{2}}{2}=\frac{\pi}{4}

⇒θ2=π2\Rightarrow \theta_{2}=\frac{\pi}{2}

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