JEE Main 2023 — Mathematics Algebra

Given,

$\alpha ,\beta ,z\in \mathbb{C}$α,β,z∈C and $\lambda >1$λ>1,

And $\sqrt{\lambda -1}$λ−1​ is the radius of the circle ${|z-\alpha |}^{2}+{|z-\beta |}^{2}=2\lambda$∣z−α∣2+∣z−β∣2=2λ

Now we know that equation of circle is given by,

${|z-{z}_{1}|}^{2}+{|z-{z}_{2}|}^{2}={|{z}_{1}-{z}_{2}|}^{2}$∣z−z1​∣2+∣z−z2​∣2=∣z1​−z2​∣2, where radius is given by $r=\frac{|{z}_{1}-{z}_{2}|}{2}$r=2∣z1​−z2​∣​,

On comparing with ${|z-\alpha |}^{2}+{|z-\beta |}^{2}=2\lambda$∣z−α∣2+∣z−β∣2=2λ, we get

Radius $\frac{|{z}_{1}-{z}_{2}|}{2}=\frac{|\alpha -\beta |}{2}$2∣z1​−z2​∣​=2∣α−β∣​ and ${|\alpha -\beta |}^{2}=2\lambda$∣α−β∣2=2λ

$$\Rightarrow \frac{|\alpha -\beta |}{2}=\sqrt{\lambda -1}(\text{given})$$⇒2∣α−β∣​=λ−1​(given)

$$\Rightarrow |\alpha -\beta |=2\sqrt{\lambda -1}$$⇒∣α−β∣=2λ−1​

$$\Rightarrow {|\alpha -\beta |}^{2}=4(\lambda -1)$$⇒∣α−β∣2=4(λ−1)

$$\Rightarrow 2\lambda =4(\lambda -1)$$⇒2λ=4(λ−1)

$$\Rightarrow \lambda =2$$⇒λ=2

Hence, ${|\alpha -\beta |}^{2}=4\Rightarrow |\alpha -\beta |=2$∣α−β∣2=4⇒∣α−β∣=2