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 i at the centre of the circle C 1 , is a circle of radius r i . If 1 = 3 , 3 = 2 3 and r 1 2 = r 2 2 + r 3 2 , then 2 is equal to

Question

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 $θ_{i}$ at the centre of the circle $C_{1}$ , is a circle of radius $r_{i}$ . 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 [2023]

Smart Achievers · Accepted Answer

(1)

If a chord of circle of radius $R$ subtends angle $θ_{i}$ at the centre $C_{1},$ then locus of the end point of this chord in a circle of radius

$r_{i} = R \cos (\frac{θ_{i}}{2})$

Given, $r_{1}^{2} = r_{2}^{2} + r_{3}^{2}$

$\Rightarrow \cos^{2} \frac{θ_{1}}{2} = \cos^{2} \frac{θ_{2}}{2} + \cos^{2} \frac{θ_{3}}{2}$

$\Rightarrow \cos^{2} \frac{π}{6} = \cos^{2} \frac{θ_{2}}{2} + \cos^{2} \frac{π}{3}$

$\Rightarrow \frac{3}{4} = \cos^{2} \frac{θ_{2}}{2} + \frac{1}{4}$

$\Rightarrow \cos^{2} \frac{θ_{2}}{2} = \frac{1}{2} \Rightarrow \cos \frac{θ_{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow \frac{θ_{2}}{2} = \frac{π}{4} \Rightarrow θ_{2} = \frac{π}{2}$

Solution

1 $\frac{π}{2}$

2 $\frac{π}{4}$

3 $\frac{π}{6}$

4 $\frac{0 - π}{4}$

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Solution

Q. The locus of the mid points of the chords of the circle C1:(x-4)2+(y-5)2=4 which subtend an angle θi at the centre of the circle C1, is a circle of radius ri. If θ1=π3,θ3=2π3 and r12=r22+r32, then θ2 is equal to [2023]

1 π2

2 π4

3 π6

4 0-π4

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1 $\frac{π}{2}$

2 $\frac{π}{4}$

3 $\frac{π}{6}$

4 $\frac{0 - π}{4}$