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Here is the problem $$$$ Show the following equation is an ellipse $$|z-1|=3-|z-2|$$ and I tried to solve it... Square both sides, $$(|z-1|)^2=(3-|z-2|)^2 \\ x^2+y^2-2x+1=9-6|z-2|+x^2+y^2-4x+4$$ Rearrange them... $$6|z-2|=12-2x \\ 3|z-2|=6-x$$ Square them again, and here is the problem i met... $$(3|z-2|)^2=(6-x)^2 \\ 9(x^2+y^2-4x+4)=36-12x+x^2 \\ $$ 36 in both sides is going to be cancelled out, it seems like I must did something wrong. I have spent over 2 hours on this but still get the same result..please help!

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So you get $8x^2+9y^2-24x=0$, which is $8\Big(x^2-3x+\frac{9}{4}\Big)-18+9y^2=0$, which is $$8\Big(x-\frac{3}{2}\Big)^2+9y^2 = 18$$ which is an ellipse.

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