Mathematics - Circle Question with Solution | TestHub

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 ${\theta }_1=\frac{\mathrm{\pi }}{3}$, ${\theta }_3=\frac{2\mathrm{\pi }}{3}$ and ${r_1}^2={r_2}^2+{r_3}^2$,

Now taking triangle $CPB$ we get,

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

Now using distance formula we get,

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

So, locus will be,

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

Now radius $r_1=2\cos \frac{\mathrm{\pi }}{6}=\sqrt{3}$, $r_2=2\cos \frac{{\mathrm{\theta }}_2}{2}$ and $r_3=2\cos \frac{\mathrm{\pi }}{3}=1$

Now using ${r_1}^2={r_2}^2+{r_3}^2$ we get,

${\left(\sqrt{3}\right)}^2={r_2}^2+1^2$

$\Rightarrow {r_2}^2=2$

$\Rightarrow 4{\cos }^2\frac{\theta }{2}=2$

$\Rightarrow \left|\cos \frac{\theta }{2}\right|=\frac{1}{\sqrt{2}}$

$\Rightarrow \theta =\frac{\mathrm{\pi }}{2}$