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Given-that-z-6-2-z-6-9i-a-Use-algebra-to-show-that-the-locus-of-z-is-a-circle-stating-its-center-and-its-radius-b-sketch-the-locus-z-on-an-argand-diagram-

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Question Number 63517 by Rio Michael last updated on 05/Jul/19
Given that ∣z−6∣=2∣z+6−9i∣, a) Use algebra to show that the locus of z is a circle, stating its center and its radius. b) sketch the locus z on an argand diagram.
Giventhat∣z−6∣=2∣z+6−9i∣,a)Usealgebratoshowthatthelocusofzisacircle,statingitscenteranditsradius.b)sketchthelocuszonanarganddiagram.
Answered by MJS last updated on 05/Jul/19
∣a−6+bi∣=2∣a+6+(b−9)i∣ (√(a^2 −12a+b^2 +36))=2(√(a^2 +12a+b^2 −18b+117)) a^2 −12a+b^2 +36=4(a^2 +12a+b^2 −18b+117) a^2 +20a+b^2 −24b+144=0 (a−p)^2 +(b−q)^2 −r^2 =0 a^2 −2pa+p^2 +b^2 −2qb+q^2 −r^2 =0 −2p=20 ⇒ p=−10 −2q=−24 ⇒ q=12 a^2 +20a+100+b^2 −24b+144−r^2 =0 a^2 +20a+b^2 −24b+144=r^2 −100 ⇒ r^2 −100=0 ⇒r=10 (a+10)^2 +(b−12)^2 −10^2 =0 ⇒ center= (((−10)),((12)) ) radius=10
∣a−6+bi∣=2∣a+6+(b−9)i∣a2−12a+b2+36=2a2+12a+b2−18b+117a2−12a+b2+36=4(a2+12a+b2−18b+117)a2+20a+b2−24b+144=0(a−p)2+(b−q)2−r2=0a2−2pa+p2+b2−2qb+q2−r2=0−2p=20⇒p=−10−2q=−24⇒q=12a2+20a+100+b2−24b+144−r2=0a2+20a+b2−24b+144=r2−100⇒r2−100=0⇒r=10(a+10)2+(b−12)2−102=0⇒center=(−1012)radius=10
Commented by Rio Michael last updated on 05/Jul/19
correct!
correct!

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