$\alpha$ is the real root and $\beta, \gamma$ are the other roots of the equation $x^3-a^3=0(a>0)$, then the…
$\alpha$ is the real root and $\beta, \gamma$ are the other roots of the equation $x^3-a^3=0(a>0)$, then the number of common points of the curves given by $|z-\beta|=\frac{\sqrt{3} a}{2}$ and $|z-\gamma|=\frac{\sqrt{3}}{2} \alpha$ is
0
2
3
1
Solution
$\because x^3-a^3=0 \Rightarrow(x-a)\left(x^2+a^2+a x\right)=0$
$\Rightarrow x=a, \frac{-a \pm \sqrt{a^2-4 a^2}}{2}$
$\begin{aligned} & \Rightarrow x=a, \frac{-a}{2} \pm i \frac{a \sqrt{3}}{2} \\ & \because x=a \text { is a real root. So, } a=\alpha \\ & \text { Now, } \beta=\frac{-a}{2}+i \frac{a \sqrt{3}}{2} \& r=\frac{-a}{2}-i \frac{a \sqrt{3}}{2} \\ & |z-\beta|=\frac{\sqrt{3} a}{2} \Rightarrow\left|x+i y+\frac{a}{2}-i \frac{a \sqrt{3}}{2}\right|^2=\left(\frac{\sqrt{3}}{2} a\right)^2 \\ & \left(x+\frac{a}{2}\right)^2+\left(y-\frac{\sqrt{3}}{2} a\right)^2=\frac{3 a^2}{4}\end{aligned}$
$x^2+y^2+a x-\sqrt{3} a y+\frac{a^2}{4}=0$ ...(i)
$\begin{aligned} & \text { Also, }|z-r|=\frac{\sqrt{3} \alpha}{2} \Rightarrow\left|x+i y+\frac{a}{2}+i \frac{a \sqrt{3}}{1}\right|^2=\left(\frac{\sqrt{3}}{2} \alpha\right)^2 \\ & \Rightarrow\left(x+\frac{a}{2}\right)^2+\left(y+\frac{a \sqrt{3}}{2}\right)^2=\frac{3}{4} \alpha^2\end{aligned}$
$\Rightarrow x^2+y^2+a x+\sqrt{3} a y+\frac{a^2}{4}=0$ ...(ii)
Equations (i) \& (ii):
$\begin{aligned} & 2 \sqrt{3} a y=0 \Rightarrow y=0 \\ & \text { From } \mathrm{eq}^{\mathrm{n}}(\mathrm{ii}) \Rightarrow x^2+0+a x+0+\frac{a^2}{4}=0 \\ & \Rightarrow\left(x+\frac{a}{2}\right)^2=0 \Rightarrow x=-\frac{a}{2} \\ & \end{aligned}$
So, only one common point $\left(-\frac{a}{2}, 0\right)$