JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 7)
The locus of the mid points of the chords of the circle $${C_1}:{(x - 4)^2} + {(y - 5)^2} = 4$$ which subtend an angle $${\theta _i}$$ at the centre of the circle $$C_1$$, is a circle of radius $$r_i$$. If $${\theta _1} = {\pi \over 3},{\theta _3} = {{2\pi } \over 3}$$ and $$r_1^2 = r_2^2 + r_3^2$$, then $${\theta _2}$$ is equal to :
$${\pi \over 2}$$
$${\pi \over 4}$$
$${{3\pi } \over 4}$$
$${\pi \over 6}$$
Explanation
_24th_January_Evening_Shift_en_7_1.png)
$\therefore \cos \left(\frac{\theta_{1}}{2}\right)=\frac{r_{i}}{2} \Rightarrow r_{i}=2 \cos \left(\frac{\theta_{i}}{2}\right)$
Given $r_{1}^{2}=r_{2}^{2}+r_{3}^{3}$
$\Rightarrow\left(\cos \left(\frac{\theta_{1}}{2}\right)\right)^{2}=\left(\cos \left(\frac{\theta_{2}}{2}\right)\right)^{2}+\left(\cos \left(\frac{\theta_{3}}{2}\right)\right)^{2}$
$\Rightarrow \frac{3}{4}=\cos ^{2}\left(\frac{\theta_{2}}{2}\right)+\frac{1}{4}$
$\Rightarrow \cos ^{2}\left(\frac{\theta_{2}}{2}\right)=\frac{1}{2}$
$\Rightarrow \frac{\theta_{2}}{2}=\frac{\pi}{4}$
$\Rightarrow \theta_{2}=\frac{\pi}{2}$
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