What is a distance function of circles outgoing from the center while shifting at constant rates towards the positive Y axis? 
I have the following problem: I want a distance equation that depending on the X and Y Coordinates will give me a distance to the circle center.
The circle center is drifting towards a certain direction, let's just say the positive Y direction, as the distance from the center increases.
The images I provided should make this clearer. The first two images show the X and the Y coordinate values respectively. The brighter the pixel the higher the value. The value 0 is assigned to black, which means that unfortunately negative values can't be displayed. For example the left side of the first iamge is completely black, but this is because the values are all negative not because X is 0.
To make visualization easier I used a rounding function, which rounds every value up to the nearest multiple of 0.1.
The third image shows the distance function of a circle when the origin is fixed at (0,0).
This is the simplest case, as it only requires the use of the pythagorean theorem.
The third image shows what I have been able to derive so far: The circle center moves towards the positive Y axis exactly by the distance between the circle center and the point at which the distance function is evaluated.
To get this function I used the fact that the distance of the point to the center and the distance of the center to the point (0,0) are equal:

(The red point is the point where the distance function is evaluated)
Using this I derived the equation below image five (the rightmost image).
One interesting thing of the equation that I derived is that it is symmetric around the X axis, which is something that I didn't intend but don't mind.
The distance function when mapped to the XY plane now looks like circles emenating from point (0,0).
Now comes my question: What would an equation for the fourth image be and how would you derive it?
As you can see when mapping the distance function it should look like the the circle origin moves along the postive Y axis such that when going in the negative Y direction the distance function increases by a constant amount.
Thank you and have a great day!
 A: Let's say the rate of vertical drift is $\alpha$, and the rate of radius change is $\beta$. The circle has centre $\vec c=(0,\alpha t)$ and radius $r=\beta t$ where $t$ is time.
A generic point $(x,y)$ on the circle has
$$\lVert(x,y)-\vec c\rVert^2=r^2$$
$$x^2+(y-\alpha t)^2=(\beta t)^2$$
$$x^2+y^2-2\alpha yt+(\alpha^2-\beta^2)t^2=0$$
Solving for $t$ with the quadratic formula, we get
$$t=\frac{2\alpha y\pm\sqrt{4\alpha^2y^2-4(\alpha^2-\beta^2)(x^2+y^2)}}{2(\alpha^2-\beta^2)}$$
$$=\frac{\alpha y\pm\sqrt{(\beta^2-\alpha^2)x^2+\beta^2y^2}}{\alpha^2-\beta^2}$$
So the radius as a function of $x$ and $y$ is
$$r=\beta t=\frac{\beta}{\alpha^2-\beta^2}\Big(\alpha y\pm\sqrt{(\beta^2-\alpha^2)x^2+\beta^2y^2}\Big)$$
It looks like you have the drift being smaller than the radius change: $\alpha<\beta$. And the radius should be positive. So we can find the proper signs in that formula:
$$r=\frac{\beta}{\beta^2-\alpha^2}\Big(-\alpha y\mp\sqrt{(\beta^2-\alpha^2)x^2+\beta^2y^2}\Big)$$
The lower sign should be taken; otherwise $r$ would be negative for some values of $x$ and $y$.
$$r=\frac{\beta}{\beta^2-\alpha^2}\Big(-\alpha y+\sqrt{(\beta^2-\alpha^2)x^2+\beta^2y^2}\Big)$$
(This is indeed always positive, because $(\beta^2-\alpha^2)x^2+(\beta^2-\alpha^2)y^2>0$ (except at the origin $x=y=0$).)
A: Following mr_e_man's brilliant answer I want to add a generalization of it. Mr_e_man's answer only allows for vertical drift however this approach can be generalized for both vertical and horizontal drift.
Instead of using radius r=βt and first calcuating t I opted for using the ratio β/α and then directly calculate the radius.
The circle has center $\vec c=(a*r,b*r)$ where a and b are the
proportionality factor between drift along the X or Y axis and the X or Y coordinate of the point being evaluated.
A generic point $(x,y)$ on the circle has
$$\lVert(x,y)-\vec c\rVert^2=r^2$$
$$(x-ar)^2+(y-br)^2=r^2$$
$$r^2(a^2+b^2-1)+r(-2xa-2yb)+x^2+y^2=0$$
When we solve this quadratic function for r we get:
$$r={a x + b y \pm \sqrt{x^2 - b^2 x^2 + 2 a b x y + y^2 - a^2 y^2}\over (-1 + a^2 + b^2)}$$
It turns out that for some reason using $-$ for $\pm $ always give positive values, so I will use that.
Therefore the general equation for the radius of a circle drifting along an arbitrary direction is:
$$r={a x + b y - \sqrt{x^2 - b^2 x^2 + 2 a b x y + y^2 - a^2 y^2}\over (-1 + a^2 + b^2)}$$
Here are some photos showing the function plotted on the XY plane:

Interestingly for $\sqrt{(a^2+b^2)}<1$ we get a picture of a Mach wave on the on side. As for what's happening on the opposite side i have no idea so please tell me if you can explain how that shape gets to be.
The avid reader will have noticed that when $a^2+b^2=1$ it will cause division through 0.
This means that we need a different function for the cases where $b=\sqrt{1-a^2}$ and $b=-\sqrt{1-a^2}$
So instead of $$(x-ar)^2+(y-b t)^2=r^2$$ we have
$$(x-ar)^2+(y-\sqrt{1-a^2}r)^2=r^2$$
and $$(x-ar)^2+(y+\sqrt{1-a^2}r)^2=r^2$$
respectively.
Solving for r give us for $b=\sqrt{1-a^2}$ $${x^2+y^2\over 2y\sqrt{1-a^2}+2xa}$$
and for $b=-\sqrt{1-a^2}$ $$-{x^2+y^2\over 2y\sqrt{1-a^2}-2xa}$$
Therefore our final equation is:
$$\require{cancel}\begin{cases}
r={a x + b y - \sqrt{x^2 - b^2 x^2 + 2 a b x y + y^2 - a^2 y^2}\over (-1 + a^2 + b^2)}   \text{  if  } a^2+b^2 \neq1 \\[2ex] 
r={x^2+y^2\over 2y\sqrt{1-a^2}+2xa}  \text{  else if  }   b=\sqrt{1-a^2} \\[2ex] 
r=-{x^2+y^2\over 2y\sqrt{1-a^2}-2xa} \text{  else  }
\end{cases}
$$
A: Circles tangential to lines through the origin can be expressed in polar coordinates $(\rho,\theta)$  with the radius vector $ OP=\rho(\theta) $.
where $d$ is the variable diameter of any circle, $\alpha $ is common inclination to x-axis, the angle between red tangent and $Ox$. Angles in the alternate segment $\theta+ \alpha $ are equal. Angle $OPQ$ in semi circle is $ 90^{\circ}$. $ OQ $ is the hypotenuse in right angled triangle $OPQ$.
$$ \rho= d \sin (\theta + \alpha )$$

This coincides with the fifth image you gave in your question as the special case $\alpha=0$. Three circle diameters are shown.
$$ \alpha = \pi/6, \quad \rho=(2,2.5,3)  $$
