I approached this by taking $y=\frac a{x+b}+c$ trying to solve the following simultaneous equations: $$\frac ab+c=t$$ $$\frac a{t+b}+c=0$$
and was only able to conclude that $b=-c$ and could not see anyway to proceed. I would appreciate a hint to continue this method or another method entirely.
I know the solution to be:
$f(x)=\frac{t(t-x)}{x+t}$
On
Consider the function $f(x)=\frac{1}{x}$. This function is symmetric with respect to the line $x=y$. We shall use this symmetry.
We can translate the graphic of any function $f$ downward by defining $g(x)=f(x)-a$, for some $a>0$. Then, we can translate the graphic of this function to the left by defining $$h(x)=g(x+a)= f(x+a)-a=\frac{1}{x+a}-a.$$ After doing that, the function $h$ still preserves the symmetry of $f$. Given $t\in\mathbb R$, we just need to find $a>0$ such that $$ \frac{1}{a}-a=t \Rightarrow a=\frac{\sqrt{t^2+4}-t}{2}. $$
There are infinitely many families of hyperbolic curves that can pass through any given two points. (in general, you need five points to uniquely determine a quadratic curve)
Your proposed family of hyperbola $\displaystyle y=\frac a{x+b}+c~$ has the two asymptotes that are vertical and horizontal. In general, the two asymptotes can have any slopes.
Of course, your hyperbola family is already more than enough.
The result $b = -c$ is correct, which means in fact you have a two-parameter family $$ y=\frac{a}{x-c}+c = \frac{a + cx - c^2}{ x - c}$$ where any real-valued $a$ and $c$ render a hyperbola with the desired property (passing through those two points).
The quoted "known solution" is just the special case when one takes $c = -t$ and $a = 2t^2$.