Question

The points of intersection of the curves $\left|z-3\right|=2$ and $\left|z\right|=2$ in an argand plane are

• A. $\frac{1}{2}(3\pm i\sqrt{7})$
• B. $\frac{1}{2}(3\pm i\sqrt{3})$
• C. $\frac{3}{2}\pm i\sqrt{\frac{7}{2}}$
• D. $\frac{7}{2}\pm i\sqrt{\frac{3}{2}})$

Answer 1

The correct option is A $\frac{1}{2}(3\pm i\sqrt{7})$

$|z-3|=2Squaringbothsides,{|z-3|}^2=4\Rightarrow Letz=x+iy\therefore {(x-3)}^2+y^2=4..........\left(1\right)\left|z\right|=2\Rightarrow {\left|z\right|}^2=4\therefore x^2+y^2=4Fromeq..........\left(1\right)Onsubstracting:9-6x=06x=9x=\frac{9}{6}=\frac{3}{2}{y}^2=4-\frac{9}{4}y=\pm \frac{\sqrt{7}}{2}\therefore x+iy=\frac{1}{2}(3\pm i\sqrt{7})$