Question Number 28219 by math solver last updated on 22/Jan/18
Commented by mrW2 last updated on 22/Jan/18
$${z}={x}+{iy} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{y}\right)^{\mathrm{2}} =\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{x}+\frac{\mathrm{1}}{\mathrm{9}}+{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{circle} \\ $$
Commented by mrW2 last updated on 22/Jan/18
$$\mid\mathrm{2}{z}−\mathrm{1}\mid=\mid{z}−\mathrm{1}\mid \\ $$$$\Rightarrow\mathrm{2}\mid{z}−\frac{\mathrm{1}}{\mathrm{2}}\mid=\mid{z}−\mathrm{1}\mid \\ $$$${distance}\:{to}\:\mathrm{1}\:{is}\:{the}\:{double}\:\left({not}\:{the}\:{same}!\right)\:{distance}\:{to}\:\mathrm{0}.\mathrm{5}. \\ $$
Commented by ajfour last updated on 22/Jan/18
$${circle} \\ $$
Commented by math solver last updated on 22/Jan/18
$${thank}\:{you}\:{sir}\:{for}\:{picking}\:{up} \\ $$$$\left.{my}\:{mistake}\::\right) \\ $$