Process for solving this system of equations I have this system of equations for which I'd like to solve for $x$,$y$, and $r$ where $a$,$b$, and $t$ are constants:
1: $0 = (x-a)^2 + (y - b)^2 - t^2$
2: $y = \dfrac{bx-rx+ar}{a}$
3: $r = \dfrac{x^2 + y^2}{2y}$
Could anyone help me get a point where one of $x$,$y$, or $r$ are defined in terms of $a$,$b$, and $t$?
I know this system is solvable in the reals (for some values of $a$,$b$, and $t$) , for example: a = 80 b = 100 t = 60, but I am unsure how WA is doing the steps of manipulation to solve.
Essentially the problem is given a circle centered at $(a,b)$ with radius $t$, find a circle which is tangent to the x-axis at $(0,0)$ and the other circle at some point $(x,y)$. 
Equation 1 is simply the equation of the original circle.
Since it is tangent to the x-axis it must be of the form $x^2 + y^2 - 2ry = 0$ where $r$ is the radius, which is where the above equation 3 comes from.
To obey tangency between the two circles, the slope from $(0,r)$ to $(a,b)$ must be the same slope as from $(0,r)$ to $(x,y)$ which is the source of equation 2.
Any insight or recommendations are greatly appreciated.
Edit: Oversight from before, tangent at (0,0)
 A: $(3)\iff (\;(y\ne 0)\land (r^2=x^2+ (y-r)^2)\;).$ Subtracting this from (1) yields a linear equation in $x, y$ (with the extra condition $y\ne 0$),which, with (2), is a system  of 2 linear equations in 2 variables.
A: I’d like to offer a perhaps simpler solution to the original problem.  
The point of tangency of the two circles lies on the line through their centers. The distance between their centers is equal to either the sum or difference of their radii. The center of a tangent circle will be at either $(x_c,y_c)=(0,r)$, $(0,-r)$ or both. After working through all of the possible sign combinations we’re left with two equations: $$(y_c\pm t)^2 = a^2+(y_c-b)^2.$$ These are linear equations for $y_c$ that give $$y_c={a^2+b^2-t^2\over2(b\pm t)}.$$ Similar triangles then yield $$x={y_c\over y_c\pm t}a \\ y=y_c+{y_c\over y_c\pm t}(b-y_c),$$ where in each of the two solutions $t$ has the same sign throughout.
A: Got it:
Substitute (3) into (2) and after some manipulation:
$$y = x * \frac{b \pm \sqrt{a^2 + b^2 - x^2}}{a+x}$$
Substitute that into (1) and after much more manipulation:
$$ x=  \frac{a^3 + ab^2 - at^2}{a^2 + b^2 \pm 2bt + t^2 }$$
Given $x$, $y$ and $r$ can be calculated by substitution.
