Geometry problem on finding the diameter of a circle $AB$ is the diameter of a circle. Tangents $AD$ and $BC$ are drawn so that $AC$ and $BD$ intersect at a point on the circle. If $|AD|=a$ and $|BC|=b$, $(a \neq b)$ then find the diameter of the circle.
 A: 
$$\tan\theta = \frac{d}{a} = \frac{b}{d} \qquad \implies \qquad d^2 = a b \qquad \implies \qquad d = \sqrt{ab} $$
A: 
Lets say AC and BD intersect at E , so at right angles as diameter subtends right angle at circumference .
Using pythagorous theorem 
$$ \displaystyle\boxed{(diameter)^2 = AE^2 + BE^2} $$
So now we should calculate AE and BE to get d 
as $$CE \times CA = CB^2 ---> 1$$ 
and $$DE \times DB = DA^2 ---> 2$$ 
and as the triangles AED and CEB are similar 
$$ \frac{AE}{EC} =  \frac{DE}{EB} = \frac{a}{b}  $$ 
$$ \frac{AC}{EC} = \frac{a+b}{b} --------> 3$$ 
$$ \frac{BD}{DE} = \frac{a+b}{a} --------> 4$$ 
substitute AC of result 3 in 1 and get $CE^2 $ and from there CE ,
Similaryly get DE
$$ CE = \frac{b*\sqrt{b}}{\sqrt{a+b}} $$
$$  DE = \frac{a*\sqrt{a}}{\sqrt{a+b}} $$
so $$AE = \frac{a}{b} \times CE $$ 
   $$\displaystyle\boxed{AE = \frac{a\sqrt{b}}{\sqrt{a+b}}}$$ 
similary 
$$BE = \frac{b}{a} \times DE $$ 
   $$\displaystyle\boxed{BE = \frac{b\sqrt{a}}{\sqrt{a+b}}}$$ 
so just substitute AE and BE in $ (diameter)^2 = AE^2 + BE^2 $and find d the diameter
$$ d^2 = \frac{a^2*b}{a+b} + \frac{b^2*a}{a+b} $$
$$ d^2 = \frac{ab*(a+b)}{a+b} $$
this implies 
$$ d = \sqrt{ab}$$
A: Consider the following diagram
$\hspace{3.2cm}$
Note that
$\triangle FBC$ is congruent to $\triangle FEC$
$\triangle FED$ is congruent to $\triangle FAD$
Thus, $\angle CFD$ is $\frac12$ of $\angle BFA$; therefore, $\angle CFD$ is a right angle.
Furthermore, $|DE|=|DA|=a$ and $|CE|=|CB|=b$
Since $\triangle CFD$ is a right triangle and $FE\perp CD$, we get that $\triangle CEF$ is similar to $\triangle FED$. This means that
$$
\frac{|CE|}{|EF|}=\frac{|EF|}{|ED|}\implies r^2=|EF|^2=|CE|\,|DE|=ab
$$
Therefore, $r=\sqrt{ab}$ .

Oops, I misread the question. Here is an answer to the actual question:
$\hspace{3.2cm}$
$\triangle BAD$ is similar to $\triangle CBA$. Therefore,
$$
\frac{|BC|}{|AB|}=\frac{|AB|}{|AD|}\implies4r^2=|AB|^2=|AD||BC|=ab
$$
Therefore, $r=\frac12\sqrt{ab}$.
A very similar answer, even though I misread the question the first time. I actually prefer the way I misread the question, so I will leave my original mis-answer.
