Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 83991 by Rio Michael last updated on 08/Mar/20

Find the locus of the points represented by the complex number ,z, such that 2∣z−3∣ = ∣z−6i∣

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{points}\:\mathrm{represented}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{number}\:,{z},\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}\mid{z}−\mathrm{3}\mid\:=\:\mid{z}−\mathrm{6i}\mid \\ $$

Commented by mathmax by abdo last updated on 08/Mar/20

let z=x+iy (e)⇔2∣x+iy−3∣=∣x+iy−6i∣ ⇒ 2(√((x−3)^2 +y^2 ))=(√(x^2 +(y−6)^2 )) ⇒4{(x−3)^2 +y^2 }=x^2 +(y−6)^2 ⇒ 4(x^2 −6x +9 +y^2 )=x^2 +y^2 −12y +36 ⇒ 4x^2 −24x +36 +4y^2 =x^2 +y^2 −12y +36 ⇒ 4x^2 −24x+4y^2 −x^2 −y^2 +12y =0 ⇒ 3x^2 +3y^2 −24x +12y =0 ⇒x^2 +y^2 −8x +4y =0 ⇒ x^2 −8x +16 −16 +y^2 +4y +4−4 =0 ⇒ (x−4)^2 +(y+2)^2 =(2(√5))^2 so the locus is circle with centre w(4,−2) and radius r=2(√5)

$${let}\:{z}={x}+{iy}\:\:\:\:\left({e}\right)\Leftrightarrow\mathrm{2}\mid{x}+{iy}−\mathrm{3}\mid=\mid{x}+{iy}−\mathrm{6}{i}\mid\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} \:+{y}^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{6}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{4}\left\{\left({x}−\mathrm{3}\right)^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right\}={x}^{\mathrm{2}} \:+\left({y}−\mathrm{6}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{6}{x}\:+\mathrm{9}\:+{y}^{\mathrm{2}} \right)={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{12}{y}\:+\mathrm{36}\:\Rightarrow \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{24}{x}\:+\mathrm{36}\:+\mathrm{4}{y}^{\mathrm{2}} ={x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} −\mathrm{12}{y}\:+\mathrm{36}\:\Rightarrow \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{4}{y}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:+\mathrm{12}{y}\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{24}{x}\:+\mathrm{12}{y}\:=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} −\mathrm{8}{x}\:+\mathrm{4}{y}\:=\mathrm{0}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}\:+\mathrm{16}\:−\mathrm{16}\:+{y}^{\mathrm{2}} \:+\mathrm{4}{y}\:+\mathrm{4}−\mathrm{4}\:=\mathrm{0}\:\Rightarrow \\ $$$$\left({x}−\mathrm{4}\right)^{\mathrm{2}} \:+\left({y}+\mathrm{2}\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \:{so}\:{the}\:{locus}\:{is}\:{circle}\:{with}\:{centre} \\ $$$${w}\left(\mathrm{4},−\mathrm{2}\right)\:{and}\:{radius}\:{r}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com