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Find-the-locus-of-the-points-represented-by-the-complex-number-z-such-that-2-z-3-z-6i-

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

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