Find the centre of the circle $z=\frac{3i-t}{2+it}$ where $t\in R$

Question

Standardising the given expression

$$\frac{t+i(6+t^2)}{4+t^2}=x+iy$$

Since it’s a circle it will be $$x^2+y^3$$ $$\implies x^2+y^2=\frac{t^2+36+t^4+12t^2}{(4+t^2)^2}$$

I can’t simplify it further.

One thing I tried was

Let $t=0$

Then $x=0$ and $y=\frac 32$ . I don’t think it’s right, just something I tried.

Answer 1

From the given $z=\frac{3i-t}{2+it}$, we have $t=\frac{3i-2z}{1+iz}$. Since $t$ is real, we have $t=\bar t$, i.e. $$\frac{3i-2z}{1+iz} = \frac{-3i-2\bar z}{1-i \bar z}$$

or,

$$|z|^2 + \frac54 i z - \frac54i \bar z +\frac32=0$$

Then, write the equation in its modular form

$$| z - \frac54 i |^2 = \frac1{16}$$

which is a circle centered at $\frac54i$.

Answer 2

The expression $x^2+y^2$ is a square distance from origin to $(x,y)$. Unless the centre of the circle coincides with the origin, this distance won't be constant. What you want to do is to find such $a,b$ independent of $t$, that $$ (x-a)^2 + (y-b)^2 = \mathrm{const} $$

Can you do this?

Another approach is to analyze functions $x(t)$ and $y(t)$. If we can find its minimal and maximal values, we can find the centre (can you guess how?) For function $x(t)$ we can notice that it is odd. And for other function: $$ y(t) = \frac{6+t^2}{4+t^2} = 1 +\frac2{4+t^2} $$

Answer 3

Hint : Try calculating \begin{eqnarray*} x^2+\left( y -\frac{5}{4} \right)^2= \cdots. \end{eqnarray*}

Answer 4

As $\lim_{t\to 0}z(t) = (0+\frac 32 i)$ and $\lim_{t\to \infty}z(t) =(0+ i)$ the center is at $(0,\frac{1}{2}(\frac 32 i+i))=(0+ \frac 54 i)$

Answer 5

$$\frac{t+i(6+t^2)}{4+t^2}=x+iy$$

$$(x, y) = \left(\dfrac{t}{4+t^2}, \dfrac{6+t^2}{4+t^2} \right)$$

$ \dfrac{dx}{dt} = \dfrac{d}{dt}\dfrac{t}{4+t^2} = -\dfrac{4t}{(4+t^2)^2}$

So the extrema for $x$ occur at $x(2)=\dfrac 14$ and at $x(-2)=-\dfrac 14$.

$ \dfrac{dy}{dt} = \dfrac{d}{dt}\dfrac{6+t^2}{4+t^2} = -\dfrac{4t}{(4+t^2)^2}$

So the extrema for $y$ occur at $y(0)=\dfrac 32$ and at $y(\pm \infty)=1$.

This implies that the center of the circle is at $\left( 0, \dfrac54 \right)$ and the radius is $\dfrac 14$.

To verify this, we compute \begin{align} \left| z - \dfrac{5i}{4} \right| &= \left| \left(\dfrac{t}{4+t^2}, \dfrac{6+t^2}{4+t^2} \right) - \left( 0, \dfrac54 \right) \right| \\ &= \left| \left(\dfrac{t}{4+t^2}, -\dfrac{t^2-4}{4(4+t^2)} \right) \right| \\ &= \dfrac{1}{16} \end{align}