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 $θ_{1}$ at the centre of the circle $C_{1}$ , is a circle of radius $r_{1}$ . If $θ_{1} = \frac{π}{3}, θ_{3} = \frac{2 π}{3}$ and $r_{1}^{2} = r_{2}^{2} + r_{3}^{2}$ , then $θ_{2}$ is equal to

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$\frac{π}{4}$

$\frac{3 π}{4}$

$\frac{π}{2}$

$\frac{π}{6}$

answer is D.

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Detailed Solution

$θ_{1} = \frac{π}{3}, θ_{2} = \frac{2 π}{3}, P (x_{1}, y_{1})$ be the mid point of the chord AB
Then $\frac{\sqrt{(x - 4)^{2} + (y - 5)^{2}}}{2} = \cos \frac{π}{6} = \frac{\sqrt{3}}{2}$
$\Rightarrow r_{1} = \sqrt{3}$
Similarly $r_{3} = 1$
Given $r_{1}^{2} = r_{2}^{2} + r_{3}^{2} \Rightarrow r_{2} = \sqrt{2}$
$∴ θ = \frac{π}{4}$