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 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.

$$\mathrm{Given}\:\mathrm{that}\:\:\mid{z}−\mathrm{6}\mid=\mathrm{2}\mid{z}+\mathrm{6}−\mathrm{9}{i}\mid, \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Use}\:\mathrm{algebra}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}, \\ $$$$\mathrm{stating}\:\mathrm{its}\:\mathrm{center}\:\mathrm{and}\:\mathrm{its}\:\mathrm{radius}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{sketch}\:\mathrm{the}\:\mathrm{locus}\:{z}\:\mathrm{on}\:\mathrm{an}\:\mathrm{argand}\:\mathrm{diagram}. \\ $$

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

$$\mid{a}−\mathrm{6}+{b}\mathrm{i}\mid=\mathrm{2}\mid{a}+\mathrm{6}+\left({b}−\mathrm{9}\right)\mathrm{i}\mid \\ $$$$\sqrt{{a}^{\mathrm{2}} −\mathrm{12}{a}+{b}^{\mathrm{2}} +\mathrm{36}}=\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{12}{a}+{b}^{\mathrm{2}} −\mathrm{18}{b}+\mathrm{117}} \\ $$$${a}^{\mathrm{2}} −\mathrm{12}{a}+{b}^{\mathrm{2}} +\mathrm{36}=\mathrm{4}\left({a}^{\mathrm{2}} +\mathrm{12}{a}+{b}^{\mathrm{2}} −\mathrm{18}{b}+\mathrm{117}\right) \\ $$$${a}^{\mathrm{2}} +\mathrm{20}{a}+{b}^{\mathrm{2}} −\mathrm{24}{b}+\mathrm{144}=\mathrm{0} \\ $$$$\left({a}−{p}\right)^{\mathrm{2}} +\left({b}−{q}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{pa}+{p}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{qb}+{q}^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$−\mathrm{2}{p}=\mathrm{20}\:\Rightarrow\:{p}=−\mathrm{10} \\ $$$$−\mathrm{2}{q}=−\mathrm{24}\:\Rightarrow\:{q}=\mathrm{12} \\ $$$${a}^{\mathrm{2}} +\mathrm{20}{a}+\mathrm{100}+{b}^{\mathrm{2}} −\mathrm{24}{b}+\mathrm{144}−{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{2}} +\mathrm{20}{a}+{b}^{\mathrm{2}} −\mathrm{24}{b}+\mathrm{144}={r}^{\mathrm{2}} −\mathrm{100}\:\Rightarrow\:{r}^{\mathrm{2}} −\mathrm{100}=\mathrm{0}\:\Rightarrow{r}=\mathrm{10} \\ $$$$\left({a}+\mathrm{10}\right)^{\mathrm{2}} +\left({b}−\mathrm{12}\right)^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{center}=\begin{pmatrix}{−\mathrm{10}}\\{\mathrm{12}}\end{pmatrix}\:\:\mathrm{radius}=\mathrm{10} \\ $$

Commented by Rio Michael last updated on 05/Jul/19

correct!

$${correct}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com