Find the point equidistant from two points and a line Given a line $l$ and two points $p_1$ and $p_2$, identify the point $v$ which is equidistant from $l$, $p_1$, and $p_2$, assuming it exists.
My idea is to: (1) identify the parabolas containing all points equidistant from each point and the line, then (2) intersect these parabolas. As $v$ is equidistant from all three and each parabola contains all points equidistant from $l$ and each point, the intersection of these parabolas must be $v$. However, I have had no luck in finding a way to compute, much less represent, these parabolas.
 A: Assuming such point $Q$ exists it must lie on the Bisector Line b of $P_1$ and $P_2$ i.e. the line through the midpoint of $P_1$ and $P_2$ and orthogonal to the line $\vec {P_1P_2}$.

Thus you can write down a parametric expression for $Q=Q(s)\in b$ and set the following equation for distances:

$$d^2(Ql)=d^2(QP_1)$$

Firstly note that, without loss of generality, we can assume that the origin coincides with the intersection point between $l$ and $b$. In fact if $b\equiv l$ solution is trivial otherwise if not we can find the intersection point and simply shift the axes (we could also rotate the axes in such way that $l$ or $b$ coincide with an axis but it should be more complex whereas shifting is trivial).
Thus, let's assume:
$P_1=(x_1,y_1), P_2=(x_2,y_2)$,
l: $ ax+by=0$,
b: $cx+dy=0$
NOTE
find c and d is trivial
Finding perpendicular bisector of the line segement joining $ (-1,4)\;\text{and}\;(3,-2)$
Parametric equation of $Q \in b$ is given by:
$Q(t)=Q(d \cdot t,-c \cdot t)$ with $t\in \mathbb{R}$
The distances are given by:
$$\text{d($Ql$)} = \frac{\left | Ax_{0} + By_{0} + C\right |}{\sqrt{A^2 + B^2} }= \frac {\left| ad\cdot t - bc\cdot t \right|}{\sqrt{a^2 + b^2}} $$
$$\text{d($QP_1$)} = \sqrt{(d\cdot t-x_1)^2 + (-c\cdot t-y_1)^2}$$
and thus
$$\frac {\left| adt - bct \right|}{\sqrt{a^2 + b^2}}=\sqrt{(d\cdot t-x_1)^2 + (-c\cdot t-y_1)^2}$$
$$\frac {\left( ad\cdot t - bc\cdot t \right)^2}{{a^2 + b^2}}=(d\cdot t-x_1)^2 + (-c\cdot t-y_1)^2$$
from which "t" and thus "Q" can be easily found.
A: Without loss of generality, $l$ is the $x$ axis, $p_1$ is at $(-a,y_1)$ and $p_2$ at $(a,y_2)$.
One of the parabolas is $$y^2=(x+a)^2+(y-y_1)^2$$ and the other
$$y^2=(x-a)^2+(y-y_2)^2.$$
After elimination of $y$,
$$y_2(x+a)^2-y_1(x-a)^2+y_1y_2(y_1-y_2)=0$$ can give you two solutions.
A: I  think you are almost there. The parabola has a property you already know.There are $two$ solutions/points for circle centers, but not one, get detected by a  direct procedure as follows:
Intersections of a properly/conveniently  placed parabola ( wlog $y=-f$ is chosen directrix) and perpendicular bisector of $P_1P_2 $ should be found.
$$ P_1:(a,b)\, ;  P_2: (0,f) ; $$ where $f$ is parabola focal length.
Bisector equation is
$$(x-a)^2+(y-b)^2= x^2+(y-f)^2$$
Simplifying 
$$ y(f-b)-ax+Q=0 ,\,where\,  Q=(a^2+b^2-f^2)/2 \tag 1 $$
Parabola equation
$$ y = \frac{x^2}{4f} \tag2 $$
Plug 2) into 1) and solve quadratic in $x$, getting two solutions, meaning two points satisfy the given condition.
The key point to realize is... there is a single parabola, not two, containing centers of these two circles on the parabola, but not one circle.
$$ x_1 = 2/ (1 - b/f)*(a + \sqrt{a^2 - Q (1 - b/f)}; \, y_1 = x_1^2/(4 f) ; $$
$$ x_2 = 2/ (1 - b/f)*(a - \sqrt{a^2 - Q (1 - b/f)}; \, y_2 = x_2^2/(4 f); $$
For the numerical example given in graph, I have taken the values $ (a,b,f)=((1,3,1)$ as numerical input.
and it results in coordinates of intersection points of circumcircle centers. By evaluation/calculation approximately they are :
$$C_1=(-4.16228,4.33114);\, C_2=(2.16228,1.16886) ; \,$$ which depicts two circumcircles as you desired.
EDIT1:
Another possibility for $l$ seems to exist as given schematically; however it is not an independent situation for the line given but reflection about line of centers.

EDIT2:
If we take $(P_1,P_2)$ as $(-c,0),(c,0)$ and the given line as $ y= m x + c, $ the equations to the (tilted double parabola ) would be simplest.
A: @anatoly has a nice answer which identifies conditions when a solution is expected. But there is an additional condition which, if not satisfied, prohibits a solution. @nominal-animal elucidates all conditions. I add an additional answer here only to show that using a different orientation of points and lines gives a compact solution for the point(s) which are equidistant to the line and the two points. It also lends itself well to constructing the solution
I prefer to let the the unit measure be the half distance between the two points and then imagine points on the x-axis at (-1,0) and (1,0) and the line having slope $m$ and y-intercept of $b$ (hence, x-intercept of $b/m$).
Using SymPy to work out the location of the point $(x,y)$ that is equidistant from the line and the points on the x-axis at $\pm 1$, we find:
>>> from sympy import Line, Point, Eq, Tuple, cse
>>> from sympy.abc import x, y, m, b, t
>>> p1 = Point(-1, 0)
>>> p2 = Point(1, 0)
>>> l = Line((0, b), slope=m)
>>> p3 = l.arbitrary_point(t)
>>> yaxis = Line(Eq(x, 0))
>>> pt = l.perpendicular_line(p3).intersection(yaxis)[0]
>>> T = Tuple(*solve(p3.distance(pt)-pt.distance(p2), t))
>>> cse(T)
([(x0, m**2 + 1), (x1, b*x0), (x2, sqrt(x0*(b - m)*(b + m))), (x3, 1/(m*x0))], [
x3*(-x1 - x2), x3*(-x1 + x2)])

Some hand simplification allows the points that are equidistant to line and the two points (here, on the x-axis) to be written as
$$
a = m^2 + 1$$$$
s = \sqrt{a \cdot (b^2 - m^2)}$$$$
e = (0, b-(a \cdot b \pm s)/m^2)
$$
Since the argument of the square root cannot be negative, we must have $b^2 >= m^2$. This is equivalent to the condition $y_1y_2>=0$ given by @anatoly. Since we cannot divide by zero, $m \neq 0$. (If the line intersects the y-axis, it cannot be a vertical line in general or collinear with the y-axis in particular). In the case of $m=0$ we have a line intersecting the y-axis at $b$ and seek a point on the y-axis $(0, y)$ that is equidistant to $(0, b)$ and $(1, 0)$ or $(-1, 0)$.
>>> solve(Point(0,y).distance((0,b))-Point(0,y).distance((1,0)), y)
[(b**2 - 1)/(2*b)]

So when the slope of the line is the same as the slope of the line connecting the two points (a slope of 0 in the orientation being used) then there is a single solution located at
$$e = (0, (b^2 - 1)/(2b))$$
And here we notice a second degenerate case: when the line passes through the two points there is no solution.
Demonstration of solution: for $m=1$ and $b=2$ we have
>>> m = 1
>>> b = 2
>>> a = m**2 + 1
>>> s = sqrt(a*(b**2 - m**2))
>>> e1, e2 = [(0, b - (a*b+sng*s)/m**2) for sgn in (-1, 1)]
>>> e1,e2
((0, -2 + sqrt(6)), (0, -sqrt(6) - 2))

These points should be equidistant from the line and either of $(-1,0)$ or $(1,0)$:
>>> line = Line((0,b), slope=m)
>>> line.distance(e1).equals(Point(e1).distance((1,0)))
True
>>> line.distance(e2).equals(Point(e2).distance((1,0)))
True

