JEE Main 2023 — Mathematics Coordinate Geometry

Given,

A line segment $AB$AB of length $\lambda$λ moves such that the points $A$A and $B$B remain on the periphery of a circle of radius $\lambda$λ,

Now taking the points on the circle of radius $\lambda$λ as B(\lambda \mathrm{cos}{\theta }_{1},\lambda \mathrm{sin}{\theta }_{1})&A(\lambda \mathrm{cos}{\theta }_{2}\lambda \mathrm{sin}{\theta }_{2}) and taking the point $P(h,k)$P(h,k) which divides the line segment in $AB$AB of length $\lambda$λ in $2:3$2:3

Now plotting the diagram we get,

Now, let $O$O be the origin and radius of circle is $\lambda$λ and $AB=\lambda$AB=λ and using distance formula we get,

$$AB=\lambda =\sqrt{{(\lambda \mathrm{cos}{\theta }_{1}-\lambda \mathrm{cos}{\theta }_{2})}^{2}+{(\lambda \mathrm{sin}{\theta }_{1}-\lambda \mathrm{sin}{\theta }_{2})}^{2}}$$AB=λ=(λcosθ1​−λcosθ2​)2+(λsinθ1​−λsinθ2​)2​

$$\Rightarrow 1=2-2\mathrm{cos}({\theta }_{1}-{\theta }_{2})$$⇒1=2−2cos(θ1​−θ2​)

$$\Rightarrow \mathrm{cos}({\theta }_{1}-{\theta }_{2})=\frac{1}{2}$$⇒cos(θ1​−θ2​)=21​

Now using section formula we get,

$h=\frac{2\lambda \mathrm{cos}{\theta }_{1}+3\lambda \mathrm{cos}{\theta }_{2}}{5}$h=52λcosθ1​+3λcosθ2​​ and $k=\frac{2\lambda \mathrm{sin}{\theta }_{1}+3\lambda \mathrm{sin}{\theta }_{2}}{5}$k=52λsinθ1​+3λsinθ2​​

Now squaring and adding above two value we get,

$${h}^{2}+{k}^{2}=\frac{{\lambda }^{2}}{25}[4+9+12(\mathrm{cos}({\theta }_{1}-{\theta }_{2}))]$$h2+k2=25λ2​[4+9+12(cos(θ1​−θ2​))]

$$\Rightarrow {h}^{2}+{k}^{2}=\frac{{\lambda }^{2}}{25}\cdot 19$$⇒h2+k2=25λ2​⋅19

Hence, Radius $=\frac{\lambda }{5}\sqrt{19}$=5λ​19​