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Question Number 19733 by Tinkutara last updated on 15/Aug/17

If ∣z + 1∣ = (√2)∣z − 1∣, then the locus described by the point z in the argand diagram is a

$$\mathrm{If}\:\mid{z}\:+\:\mathrm{1}\mid\:=\:\sqrt{\mathrm{2}}\mid{z}\:−\:\mathrm{1}\mid,\:\mathrm{then}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{described}\:\mathrm{by}\:\mathrm{the}\:\mathrm{point}\:{z}\:\mathrm{in}\:\mathrm{the}\:\mathrm{argand} \\ $$$$\mathrm{diagram}\:\mathrm{is}\:\mathrm{a} \\ $$

Answered by ajfour last updated on 15/Aug/17

let z=x+iy ∣(x+1)+iy∣^2 =2∣(x−1)+iy∣^2 ⇒ x^2 +2x+1+y^2 =2x^2 −4x+2+2y^2 or x^2 −6x+y^2 +1=0 (x−3)^2 +y^2 =(2(√2))^2 or ∣z−3∣=2(√2) .

$$\mathrm{let}\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\mid\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{iy}\mid^{\mathrm{2}} =\mathrm{2}\mid\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{iy}\mid^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}+\mathrm{y}^{\mathrm{2}} =\mathrm{2x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{2}+\mathrm{2y}^{\mathrm{2}} \\ $$$$\mathrm{or}\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{y}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{or}\:\:\:\:\:\:\:\:\mid\mathrm{z}−\mathrm{3}\mid=\mathrm{2}\sqrt{\mathrm{2}}\:. \\ $$

Commented by Tinkutara last updated on 15/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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