For the two equations you have, expand the squares on the left sides. You'll have $s^2$ and $r^2$ on the left side in both equations. Subtracting the two equations gives an equation that is linear in $s$ and $r$. Solve the linear equation for one variable and substitute into one of your original two equations to solve for the other variable. I believe that there are two solutions (by symmetry over the line through the known center and the known circumference point).
edit Let me redefine your variables, then go through some of the algebra. I'll continue to call the known center $(h,k)$ and the radius of that circle $r_1$ and the radius of the other circle $r_2$; I'll call the known point on the circumference of the other circle $(a,b)$, the desired point of tangency $(c,d)$, and the unknown center of the other circle $(m,n)$. The system of equations you had, written in my variables, is:
$$\begin{align}
(a-m)^2+(b-n)^2&=r_2^2
\\
(m-h)^2+(n-k)^2&=(r_1+r_2)^2
\end{align}$$
We want to solve for $(m,n)$. Expand the squares (use $(x+y)^2=x^2+2xy+y^2$) on the left sides of both equations:
$$\begin{align}
a^2-2am+m^2+b^2-2bn+n^2&=r_2^2
\\
m^2-2mh+h^2+n^2-2nk+k^2&=(r_1+r_2)^2
\end{align}$$
Subtract the two equations (I'm subtracting the first from the second):
$$(m^2-2mh+h^2+n^2-2nk+k^2)-(a^2-2am+m^2+b^2-2bn+n^2)=(r_1+r_2)^2-r_2^2$$
$$m(2a-2h)+n(2b-2k)+(h^2+k^2-a^2-b^2)=r_1^2+2r_1r_2$$
I've grouped the terms on the left side into $m$ times some constant, $n$ times some constant, and a constant term (only $m$ and $n$ are variables here); I'll continue to group terms in this way. Solve for one of the variables ($m$):
$$\begin{align}
m(2a-2h)
&=r_1^2+2r_1r_2-n(2b-2k)-(h^2+k^2-a^2-b^2)
\\
&=(r_1^2+2r_1r_2-h^2-k^2+a^2+b^2)-n(2b-2k)
\end{align}$$
$$m=\frac{(r_1^2+2r_1r_2-h^2-k^2+a^2+b^2)-n(2b-2k)}{2a-2h}$$
Now, substitute this expression for $m$ in one of the two original equations (I'll use the first):
$$(a-\frac{(r_1^2+2r_1r_2-h^2-k^2+a^2+b^2)-n(2b-2k)}{2a-2h})^2+(b-n)^2=r_2^2$$
This is a (very messy) quadratic equation in $n$ (it can be written in the form $()n^2+()n+()=0$; $n$ is the only variable; the rest are constants). While it's possible to continue to solve by hand, the symbolic manipulation is messy (though I can fill in more detail if necessary). The two solutions for $n$ (note the $\pm$) that result are:
$n=\scriptstyle\frac{a^2 (b+k)-2 a h (b+k)+h^2 (b+k)+(b-k) \left(b^2-k^2+r_1^2+2 r_1r_2\right)\pm\sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left(a^2-2 a h+(b-k)^2+(h-r_1-2 r_2) (h+r_1+2 r_2)\right)}}{2 \left((a-h)^2+(b-k)^2\right)}$
The next step would be to take this, put it back into the expression for $m$ to get two values of $m$. Here are the results I get using Mathematica:
$(m,n)=$
$\begin{matrix}\scriptstyle{\left(
\frac{1}{2 (a-h) \left((a-h)^2+(b-k)^2\right)}
\left(\scriptstyle{\begin{align}
a^4&-2 a^3 h+2 a h^3-h^2 \left(h^2+(b-k)^2\right)+a^2 (b-k)^2+(a-h)^2 r_1 \left(r_1+2 r_2\right)\\
&+b \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}\\
&-k \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}
\end{align}}\right)\right.},\\ \quad\quad\scriptstyle{\left.
\frac{\left((a-h)^2+(b-k)^2\right) (b+k)+(b-k) r_1 \left(r_1+2 r_2\right)-\sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}}{2 \left((a-h)^2+(b-k)^2\right)}\right)}\end{matrix}$
or $(m,n)=$
$\begin{matrix}\scriptstyle{\left(
\frac{1}{2 (a-h) \left((a-h)^2+(b-k)^2\right)}
\left(\scriptstyle{\begin{align}
a^4&-2 a^3 h+2 a h^3-h^2 \left(h^2+(b-k)^2\right)+a^2 (b-k)^2+(a-h)^2 r_1 \left(r_1+2 r_2\right)\\
&-b \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}\\
&+k \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}
\end{align}}\right)\right.},\\ \quad\quad\scriptstyle{\left.
\frac{\left((a-h)^2+(b-k)^2\right) (b+k)+(b-k) r_1 \left(r_1+2 r_2\right)+\sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}}{2 \left((a-h)^2+(b-k)^2\right)}\right)}\end{matrix}$
Since we know the two circles are tangent at the point $(c,d)$, the point $(c,d)$ is on the line connecting the centers of the circles and $\frac{r_1}{r_1+r_2}$ of the way from $(h,k)$ to $(m,n)$, so $$(c,d)=\left(\frac{r_2}{r_1+r_2}h+\frac{r_1}{r_1+r_2}m,\frac{r_2}{r_1+r_2}k+\frac{r_1}{r_1+r_2}n\right).$$
Mathematica isn't generating any nice simplification of that expression with $m$ and $n$ substituted in, so I'll stop here.
The radius of the large circle isn’t an independent variable. It’s equal to the distance from the circle’s center to either of the chord endpoints, so it’s a simple function of $t$.
If you subtract the equation of the small circle from that of the large one, you’ll get a linear equation, that of their radical axis. If the two circles are tangent, then this line is also tangent to them both. Applying this constraint should give you an ugly, but straightforward to solve quadratic equation in $t$.
I would’ve parameterized the family of large circles a bit differently. All of the circles that share the given chord have equations of the form $f(x,y)+\lambda g(x,y)=0$, where $f(x,y)=0$ is the equation of the circle that has the chord as its diameter and $g(x,y)=0$ is the equation of the line that contains the chord. Proceeding as I described above again leads to a quadratic equation in $\lambda$, and the radii of the resulting circles are easily extracted from the equations that result from solving for $\lambda$ and substituting.
Best Answer
The circle's center would be at the intersection of a line L and a parabola P.
The line L comes from being equidistant from your line 1 and line 2. If these intersect, take the (two) angle bisectors through that point of intersection as line L. If line 1 and line 2 are parallel, take line L to be the parallel line halfway between them.
For parabola P take your known point (xp,yp) that the circle passes through as the focus and say line 1 (or line 2 if convenient) as your directrix. That is, the center of the circle will be equidistant from the known point (xp,yp) and the point of tangency to the directrix, which amounts to affirming the eccentricity 1 of a parabola.
Note that unless the directrix is parallel to an axis, the parabola will be in "general position", which means the equation will be messier than necessary. It probably pays to translate and rotate the coordinates so that the directrix is parallel to (say) the x-axis, and for that matter so that the focus and directrix are equidistant from the origin (so that the parabola will pass through the origin and have a simple form $y = ax^2$).
Added: Since the known point (xp,yp) must lie on the same side of lines 1 and 2 as the circle and its center, choose line L through the intersection of lines 1 and 2 so that it also extends into that portion of the plane.