Area of Kite Formed By Tangents to Circle and Radii

What is the area of the kite formed by tangents to a circle and radii to those tangents?

The diagram shows such a kite The area is
$A= \frac{1}{2}( AO \times BC)$

$BC= \sqrt{(3-1)^2+(4-1)^2}= \sqrt{13}$
.
To find the other diagonal we use that triangles ABC and BQO are similar triangles, so
$\frac{BQ}{BO}=\frac{AB}{AO}$
.
$AB=\sqrt{AO^2-BO^2}= \sqrt{13-2^2}=3$
, then
$\frac{BQ}{BO}=\frac{AB}{AO} \rightarrow BQ= \frac{AB}{AO} BO=\frac{3}{\sqrt{13}} \times 2 = \frac{6}{\sqrt{13}}$
.
Then
$BC=\frac{12}{\sqrt{13}}$
.
The area of the kite is
$\frac{1}{2} \times \sqrt{13} \times \frac{12}{\sqrt{13}}=6$
.