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

$Find the locus of the points represented by$ $the complex number, z, such that$ $2 ∣ z - 3 ∣ = ∣ z - 6 i ∣$
Commented by mathmax by abdo last updated on 08/Mar/20
$let z=x+iy (e)⇔2∣x+iy−3∣=∣x+iy−6i∣ ⇒ 2(√((x−3)^2 +y^2 ))=(√(x^2 +(y−6)^2 )) ⇒4{(x−3)^2 +y^2 }=x^2 +(y−6)^2 ⇒ 4(x^2 −6x +9 +y^2 )=x^2 +y^2 −12y +36 ⇒ 4x^2 −24x +36 +4y^2 =x^2 +y^2 −12y +36 ⇒ 4x^2 −24x+4y^2 −x^2 −y^2 +12y =0 ⇒ 3x^2 +3y^2 −24x +12y =0 ⇒x^2 +y^2 −8x +4y =0 ⇒ x^2 −8x +16 −16 +y^2 +4y +4−4 =0 ⇒ (x−4)^2 +(y+2)^2 =(2(√5))^2 so the locus is circle with centre w(4,−2) and radius r=2(√5)$
$l e t z = x + i y (e) \Leftrightarrow 2 ∣ x + i y - 3 ∣=∣ x + i y - 6 i ∣ \Rightarrow$ $2 \sqrt{{(x - 3)}^{2} + y^{2}} = \sqrt{x^{2} + {(y - 6)}^{2}} \Rightarrow 4 {{(x - 3)}^{2} + y^{2}} = x^{2} + {(y - 6)}^{2} \Rightarrow$ $4 (x^{2} - 6 x + 9 + y^{2}) = x^{2} + y^{2} - 12 y + 36 \Rightarrow$ $4 x^{2} - 24 x + 36 + 4 y^{2} = x^{2} + y^{2} - 12 y + 36 \Rightarrow$ $4 x^{2} - 24 x + 4 y^{2} - x^{2} - y^{2} + 12 y = 0 \Rightarrow$ $3 x^{2} + 3 y^{2} - 24 x + 12 y = 0 \Rightarrow x^{2} + y^{2} - 8 x + 4 y = 0 \Rightarrow$ $x^{2} - 8 x + 16 - 16 + y^{2} + 4 y + 4 - 4 = 0 \Rightarrow$ ${(x - 4)}^{2} + {(y + 2)}^{2} = {(2 \sqrt{5})}^{2} s o t h e l o c u s i s c i r c l e w i t h c e n t r e$ $w (4, - 2) a n d r a d i u s r = 2 \sqrt{5}$
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