The locus of the midpoints of the chords of the circle C_1:(x-4)^2+(y-5)^2=4, which subten…

Question

The locus of the midpoints of the chords of the circle $C_1:(x-4)^2+(y-5)^2=4$, which subtend an angle $	heta_i$ at the centre of $C_1$, is a circle of radius $r_i$. If $	heta_1=\dfrac{\pi}{3}$, $	heta_3=\dfrac{2\pi}{3}$, and $r_1^2=r_2^2+r_3^2$, then $	heta_2$ is equal to:

Accepted Answer

$	herefore \cos \left(\frac{	heta_{1}}{2}ight)=\frac{r_{i}}{2} \Rightarrow r_{i}=2 \cos \left(\frac{	heta_{i}}{2}ight)$

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

$\Rightarrow\left(\cos \left(\frac{	heta_{1}}{2}ight)ight)^{2}=\left(\cos \left(\frac{	heta_{2}}{2}ight)ight)^{2}+\left(\cos \left(\frac{	heta_{3}}{2}ight)ight)^{2}$

$\Rightarrow \frac{3}{4}=\cos ^{2}\left(\frac{	heta_{2}}{2}ight)+\frac{1}{4}$

$\Rightarrow \cos ^{2}\left(\frac{	heta_{2}}{2}ight)=\frac{1}{2}$

$\Rightarrow \frac{	heta_{2}}{2}=\frac{\pi}{4}$

$\Rightarrow 	heta_{2}=\frac{\pi}{2}$

Solutions · 1