Given the set $|z-i|= k|z-2i|$ when $k>0$
I don't understand how to sketch this set.
Help me please. Thank you.
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From $|z-i|= k|z-2i|$
I have $\sqrt{x^2+(y-1)^2} = k\sqrt {x^2+(y-2)^2}$
I have $ x^2+(y-1)^2 = k^2(x^2+(y-2)^2)$
Thus, $x^2+y^2-2y+1 = k^2(x^2 +y^2-4y+4)$
Suppose $k=\frac{1}{2}$ , I have $x^2+(y-\frac{2}{3})^2 = \frac{4}{9}$
Suppose $k=1$ , I have $y=\frac{3}{2}$
Suppose $k=2$ , I have $x^2+(y-\frac{7}{3})^2 =\frac{19}{6} $
Suppose $k=3$ , I have $x^2+(y-\frac{17}{8})^2 = \frac{9}{64}$
Now, I think when I suppose $k\geq{3}$ I have circles small than $k=2$
Can I sketch this set from equation $x^2+(y-\frac{7}{3})^2 =\frac{19}{6} $ ?
Continue with $x^2+y^2-2y+1 = k^2(x^2 +y^2-4y+4)$ to obtain the general expression,
$$x^2+\left(y-\frac{2k^2-1}{k^2-1}\right)^2 = \left(\frac k{k^2-1}\right)^2$$
So, for any $k>0$ and $k\ne1$, the set is of a circle with center $(0,\frac{2k^2-1}{k^2-1})$ and radius $\frac k{k^2-1}$. For example, if $k=2$, the circle becomes,
$$x^2+\left(y-\frac{7}{3}\right)^2 = \left(\frac 2{3}\right)^2$$
which is sketched here, showing that the distance from $z$ to (0,i) is twice as that from $z$ to (0,2i).