Interesting geometry problem (3 circle puzzle)

The key observation is that the triangle formed by the three centres of the circles has a right-angle at the centre of the larger black circle (note the marked $45^{\circ}$ angle).

Let the radius of the blue circle be $r$ , the larger black circle $a$ and the smaller black circle $b$ . Then we get the following equations:

$\begin{aligned} r&=2a+6 \\ r&=2b+12 \\ (a+6)^2+(a+b)^2 &=(b+12)^2 \end{aligned}$

Substituting for $a$ and $b$ in the third equation, we get the quadratic $r^2 - 21 r + 54 = 0$ ; solving and discarding the smaller root (which makes $a$ and $b$ negative), we get $r=\boxed{18}$ .

firstly if there are two circles with centres $A,C$ and intersects at only point $B$ then $B\in [AC]$

because these two circles tangent on point $B$ will be the same line

$\angle ABD = 90^\circ$ , $\angle CBD = 90^\circ$ and sum of them is $180^\circ$ which shows that $A$ , $B$ , $C$ are linear

let big circles radius is $R$ and little ones $r$

$2R+6=2r+12 \Rightarrow R=r+3$

draw the line segment between black circles centres the little triangle will be $45^\circ-45^\circ-90^\circ$ triangle

on the big right triangle : $(r+3+r)^{2}+(r+3+6)^{2}=(12+r)^{2}$

$\Rightarrow 2r^{2}+3r-27=0$

$(2r+9)(r-3)=0$ Hence $r=3$ . blue circles radius is $12+2r=18$