JEE Main 2017 (Online) 8th AprilQuestion 1
Let $z\in C$, the set of complex numbers. Then the equation $2|z+3i|-|z-i|=0$ represents :
Complex Numbers
Solution
$4(x^2+(y+3)^2)=x^2+(y-1)^2$ gives $3x^2+3y^2+26y+35=0$: centre $(0,-\frac{13}{3})$, $r^2=\frac{169}{9}-\frac{35}{3}=\frac{64}{9}$, a circle of radius $\frac83$.