Show Posts
1
Quiz-1 / Re: Q1: TUT 5201
« on: September 28, 2018, 05:49:13 PM »
Let $z=x+yi $
Then the equation can be rewritten as:
\begin{gather*}
|(x+yi)+1|^2+2|x+yi|^2=|(x+yi)-1|^2\\
|(x+1)+yi|^2+2|x+yi|^2=|(x-1)+yi|^2\\
(x+1)^2+y^2+2x^2+2y^2=(x-1)^2+y^2\\
3x^2+2x+3y^2+1=x^2-2x+y^2+1\\
2x^2+2y^2+4x=0\\
(x+1)^2+y^2=1
\end{gather*}
Therefore, the locus is circle centered at $(-1,0)$ with radius $1$
Fixed formatting. V.I.
Then the equation can be rewritten as:
\begin{gather*}
|(x+yi)+1|^2+2|x+yi|^2=|(x+yi)-1|^2\\
|(x+1)+yi|^2+2|x+yi|^2=|(x-1)+yi|^2\\
(x+1)^2+y^2+2x^2+2y^2=(x-1)^2+y^2\\
3x^2+2x+3y^2+1=x^2-2x+y^2+1\\
2x^2+2y^2+4x=0\\
(x+1)^2+y^2=1
\end{gather*}
Therefore, the locus is circle centered at $(-1,0)$ with radius $1$
Fixed formatting. V.I.