Solve this following


If the equation $a|z|^{2}+\overline{\bar{\alpha} z+\alpha \bar{z}}+d=0$ represents a circle where a,d are real constants then which of the following condition is correct?


  1. $|\alpha|^{2}-\mathrm{ad} \neq 0$

  2. $|\alpha|^{2}-a d>0$ and $a \in R-\{0\}$

  3. $|\alpha|^{2}-a d \geq 0$ and $a \in R$

  4. $\alpha=0, a, d \in R^{+}$

Correct Option: , 2


az $\bar{z}+\alpha \bar{z}+\bar{\alpha} z+d=0 \rightarrow$ Circle

centre $=\frac{-\alpha}{a} \quad 2=\sqrt{\frac{\alpha \bar{\alpha}}{a^{2}}-\frac{d}{a}}=\sqrt{\frac{\alpha \bar{\alpha}-a d}{a^{2}}}$

So $|\alpha|^{2}-\mathrm{ad}>0 \& \mathrm{a} \in \mathrm{R}-\{0\}$


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