Finding *two* equations of circles given 2 common points and a common tangent. Suppose we're given 2 points on a circle, $A(1,3)$ and $B(2,4)$, and its tangent being the $y$-axis. We're asked to find the equation of the circle.
If I use the standard procedure $$(x_n-h)^2+(y_n-k)^2=r^2$$ and $$r=\frac{|ax_n+by_n+d|}{\sqrt{a^2+b^2}}$$ I will get only one equation of a circle but we know that, in this case, there should be 2 possible circles. How do we get the other equation? 
A quick sketch to show the possibility of two circles given the set of information.

 A: The circle with equation $(x-h)^2 + (y-k)^2 = r^2$ is tangent to the $y$ axis if and only if  $h=\pm r$.  With $h = r$, plugging in $x=1, y=3$ and $x=2,y=4$ gives  two equations that simplify to 
$$ \eqalign{{k}^{2}-6\,k-2\,r+10 &= 0\cr{k}^{2}-8\,k-4\,r+20  &= 0}$$
and eliminating $r$ we get
$ k^2-4k = 0$.  Thus either $k=0$ (in which case $r=5$) or $k=4$ (and $r=1$).
With $h=-r$, we get $k=0, r=-5$ or $k=4, r=-1$, but negative $r$ is not allowed.

A: Your “standard procedure” should’ve found both solutions, but, at least as you’ve written the equations in your question, you measured the distance to the tangent line from the wrong point. The equation of a circle tangent to the line $ax+by+c=0$ is $$(x-h)^2+(y-k)^2={(ah+bk+c)^2 \over a^2+b^2}.\tag 1$$ In this question the tangent line is $x=0$. Substituting for $a$, $b$ and $c$ on the right-hand side and plugging the coordinates of the two known points into the left produces the system $$(1-h)^2+(3-k)^2=h^2 \\ (2-h)^2+(4-k)^2=h^2,$$ which has two solutions. (Start by subtracting one equation from the other to eliminate the squared terms, solve for one variable in terms of the other, and back-substitute.)
Here’s a somewhat different approach that requires solving a single quadratic equation instead: An equation of the circle with diameter $\overline{AB}$ is $(x-1)(x-2)+(y-3)(y-4)=0$, while the line through the two points is $x-y+2=0$. Combining these two equations gives the one-parameter family of circles through the two points: $$(x-1)(x-2)+(y-3)(y-4)+\lambda(x-y+2)=0. \tag 2$$ (If you consider the line as part of this family, then use a convex combination of the two equations instead.)  
A circle in this family is tangent to the line $x=0$ iff it has exactly one intersection with this line. Setting $x=0$ in (1) and simplifying produces $$y^2-(\lambda+7)y+2\lambda+14=0,$$ which has a single solution for $y$ when $\lambda^2+6\lambda-7=0$. Solve for $\lambda$ and substitute into (2).
A: Note that $AB$ intersects the $y$-axis at $C=(0,2)$. Note that $AC\cdot BC = 4$, so by power of a point, the tangency point $P$ must satisfy $CP^2 = 4$, i.e. $P$ is $(0,0)$ or $(0,4)$. Equations for the circles can now easily be computed.
