Finding a point a certain distance away from 2 points I need to find a point that is a certain distance away from two known points.

Where $P_1, P_2, L_2$ and $L_1$ are all defined and that is all that is known. How do I find $P_3?$
Kind Regards.
 A: Taking Brondahl's answer one step further,
if we expand the equations,
we get
$x^2-2xP_{1X}+P_{1X}^2 + y^2-2yP_{1Y}+P_{1Y}^2 = L_1^2$
and
$x^2-2xP_{2X}+P_{2X}^2 + y^2-2yP_{2Y}+P_{2Y}^2 = L_2^2$
.
Subtracting these,
to get rid of the quadratic terms,
we get
$2x(P_{2X}-P_{1X})+2y(P_{2Y}-P_{1Y})
+P_{1X}^2-P_{2X}^2+P_{1Y}^2-P_{2Y}^2
=L_1^2-L_2^2
$.
This is a linear equation
in $x$ and $y$
of the form
$ax+by = c$.
If $b \ne 0$,
we can solve this for $y$
in terms of $x$
as $y = \frac{c-ax}{b}$.
If $a \ne 0$,
we can solve this for $x$
in terms of $y$
as $x = \frac{c-by}{a}$.
We can not have both
$a=0$ and $b=0$, 
for this would mean that
$P_1$ and $P_2$ are the same.
Putting in the expression
for $x$ or $y$
in one of the quadratics,
we get a quadratic equation
for
$y$ or $x$.
Solving this equation,
we generally get two solutions.
This is reasonable,
since the geometrical solution
involves the intersection of two circles,
which generally intersect,
if they intersect at all,
in two points.
Note that
if the distance from $P_1$
to $P_2$ is greater than
$L_1+L_2$,
there will be no real solution,
since the circles can not intersect.
You can show this algebraically.
A: Draw a circle with center $P_1$ and radius $l_1$ and a circle with center $P_2$ and radius $l_2$, their intersection(s) is what you want.
A: Salomo's answer is correct if you're looking for a geometric, pen and paper solution.
For an algebraic solution (as suggested by your tags), you essentially do the same conceptual thing.
To translate the geometric approach literally, the "circles" become 2 circle equations (each of the form $k = (x+a)^2 + (y+b)^2$) And the intersection becomes the solution to the simultaneous equations representing both equations being simultaneously true.
So solve the following for x and y to find $P_{3X}$ & $P_{3Y}$:


*

*$(x-P_{1X})^2 + (y-P_{1Y})^2 = L_1^2$

*$(x-P_{2X})^2 + (y-P_{2Y})^2 = L_2^2$
