The locus of the middle points of the chords of the circle C _ 1 :...

Given,

${C}_{1}: {\left(x - 4\right)}^{2} + {\left(y - 5\right)}^{2} = 4$ which subtend an angle ${\theta}_{i}$ at the centre of the circle ${C}_{i}$, is a circle of radius ${r}_{i}$,

Now plotting the diagram we get,

Now given If ${\theta}_{1} = \dfrac{\pi}{3}$, ${\theta}_{3} = \dfrac{2\pi}{3}$ and ${{r}_{1}}^{2} = {{r}_{2}}^{2} + {{r}_{3}}^{2}$,

Now taking triangle $CPB$ we get,

$cos\dfrac{\theta}{2} = \dfrac{PC}{2}$, so $PC = 2cos\dfrac{\theta}{2}$

Now using distance formula we get,

${\left(h - 4\right)}^{2} + {\left(k - 5\right)}^{2} = 4\cos^{2}\dfrac{\theta}{2}$

So, locus will be,

${\left(x - 4\right)}^{2} + {\left(y - 5\right)}^{2} = 4\cos^{2}\dfrac{\theta}{2}$

Now radius ${r}_{1} = 2cos\dfrac{\pi}{6} = \sqrt{3}$, ${r}_{2} = 2cos\dfrac{{\theta}_{2}}{2}$ and ${r}_{3} = 2cos\dfrac{\pi}{3} = 1$

Now using ${{r}_{1}}^{2} = {{r}_{2}}^{2} + {{r}_{3}}^{2}$ we get,

${\left(\sqrt{3}\right)}^{2} = {{r}_{2}}^{2} + {1}^{2}$

$⇒ {{r}_{2}}^{2} = 2$

$⇒ 4\cos^{2}\dfrac{\theta}{2} = 2$

$⇒ \left|\cos\dfrac{\theta}{2}\right| = \dfrac{1}{\sqrt{2}}$

$⇒ \theta = \dfrac{\pi}{2}$