Q:

Enlarge 3 Circles about the same factor to find the Intersect

I currently have 3 circles that not intersect at all.
Like this:

Now i would like enlarge the circles about the same factor to find the intersection of this three circles. 
I have tried following some other intersecting posts and maths but haven't managed to fully understand the principles behind them.
Hope somebody can give me a hint or explain me if it's working.
thanks

A:

Our first task is to, given two of the circles, find the places where they will intersect as we grow them.  This will be a circle (which I'll call a locus circle), though I won't prove it here.
WLOG we can assume that the circles are 1 unit apart, and that the larger of the two has radius $a$ and the smaller has radius $b$.  There are then two scale factors that are easy to calculate and give known locations:
$$aj + bj = 1$$
which describes a point between the two circle centers, and 
$$ak = bk + 1$$
which describes a point somewhere past the center of the smaller circle.  Solving for their distances from the center of circle $A$, we get
$$aj=\frac{a}{a+b}$$
$$ak=\frac{a}{a-b}$$
Halfway between these is the center of the locus circle.
$$t=a\frac{k+j}{2}=\frac{a^2}{a^2-b^2}$$
And half the distance between them is the radius of the locus circle.
$$r=a\frac{k-j}{2}=\frac{ab}{a^2-b^2}$$
Now that we know what the circle looks like, let's put it back into the real space of our circles.
The center of the circle is based on the centers of the original circles, which we'll call $A$ and $B$.
$$P = (1-t)A+tB$$
The radius is a distance, and we need the original distance between $A$ and $B$: $|A-B|$.
$$p = r|A-B|$$
This process so far -- using the formulas for $t$, $r$, $P$, and $p$ -- gets you one of two locus circles.  Now we need to get another, which works the same way; just use $A$ and $C$ instead, to create $Q$ and $q$.
Once we have the two locus circles, we can find their intersections (of which, hopefully, there are two).  We'll call those $X$ and $Y$.
Your growth factor then is $$F = \frac{|X-A|}{a}$$
Or maybe it's $Y-A$ instead; either of the two is a location where all three circles can meet.

