JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
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 \(\theta_1=\frac{\pi}{3}, \theta_3=\frac{2 \pi}{3}\) and \(r_1^2=r_2^2+r_3^2\), then \(\theta_2\) is equal to
- A \(\frac{\pi}{4}\)
- B \(\frac{3 \pi}{4}\)
- C \(\frac{\pi}{6}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
In \(\triangle C P B\) \(\cos \frac{\theta}{2}=\frac{P C}{2} \Rightarrow P C=2 \cos \frac{\theta}{2}\) \(\Rightarrow(h-4)^2+(k-5)^2=4 \cos ^2 \frac{\theta}{2}\) \(\text { Now }(x-4)^2+(y-5)^2=\left(2 \cos \frac{\theta}{2}\right)^2\)…
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