JEE Main 2023 — Mathematics Coordinate Geometry

Given,

${C}_{1}:{(x-4)}^{2}+{(y-5)}^{2}=4$C1​:(x−4)2+(y−5)2=4 which subtend an angle ${\theta }_{i}$θi​ at the centre of the circle ${C}_{i}$Ci​, is a circle of radius ${r}_{i}$ri​,

Now plotting the diagram we get,

Now given If ${\theta }_{1}=\frac{\pi }{3}$θ1​=3π​, ${\theta }_{3}=\frac{2\pi }{3}$θ3​=32π​ and ${{r}_{1}}^{2}={{r}_{2}}^{2}+{{r}_{3}}^{2}$r1​2=r2​2+r3​2,

Now taking triangle $CPB$CPB we get,

$\mathrm{cos}\frac{\theta }{2}=\frac{PC}{2}$cos2θ​=2PC​, so $PC=2\mathrm{cos}\frac{\theta }{2}$PC=2cos2θ​

Now using distance formula we get,

$${(h-4)}^{2}+{(k-5)}^{2}=4{\mathrm{cos}}^{2}\frac{\theta }{2}$$(h−4)2+(k−5)2=4cos22θ​

So, locus will be,

$${(x-4)}^{2}+{(y-5)}^{2}=4{\mathrm{cos}}^{2}\frac{\theta }{2}$$(x−4)2+(y−5)2=4cos22θ​

Now radius ${r}_{1}=2\mathrm{cos}\frac{\pi }{6}=\sqrt{3}$r1​=2cos6π​=3​, ${r}_{2}=2\mathrm{cos}\frac{{\theta }_{2}}{2}$r2​=2cos2θ2​​ and ${r}_{3}=2\mathrm{cos}\frac{\pi }{3}=1$r3​=2cos3π​=1

Now using ${{r}_{1}}^{2}={{r}_{2}}^{2}+{{r}_{3}}^{2}$r1​2=r2​2+r3​2 we get,

$${(\sqrt{3})}^{2}={{r}_{2}}^{2}+{1}^{2}$$(3​)2=r2​2+12

$$\Rightarrow {{r}_{2}}^{2}=2$$⇒r2​2=2

$$\Rightarrow 4{\mathrm{cos}}^{2}\frac{\theta }{2}=2$$⇒4cos22θ​=2

$$\Rightarrow |\mathrm{cos}\frac{\theta }{2}|=\frac{1}{\sqrt{2}}$$⇒∣cos2θ​∣=2​1​

$$\Rightarrow \theta =\frac{\pi }{2}$$⇒θ=2π​