## Area of Kite Formed By Tangents to Circle and Radii

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

\[\frac{1}{2} \times \sqrt{13} \times \frac{12}{\sqrt{13}}=6\]

.