$AB\text{ is a chord of the circle }{x}^2+y^2=25.\text{ The tangents to the circle at }A\text{ and }B\text{ intersect }$$\text{ at }C\text{ . If }(2,3)\text{ is the midpoint of }AB,\text{ then the area of quadrilateral }OACB\text{ (Where }O\text{ is }$$\text{ origin) is }$

$\begin{array}{l}\text{ Here }OB=OA=5\text{ ( radius of circle) }\\ OP=\sqrt{(2-0{)}^2+(3-0{)}^2}=\sqrt{13}\\ \text{ Let }\mathrm{\angle }OBP=\theta ,\text{ then }\mathrm{\angle }PCB=\theta \\ \text{ From }\mathrm{\Delta }OPB,\cos (90-\theta )=\frac{OP}{OB}\\ =\frac{\sqrt{13}}{5}\\ \Rightarrow \sin \theta =\frac{\sqrt{13}}{5}\\ \Rightarrow \cot \theta =\frac{2\sqrt{3}}{\sqrt{13}}\end{array}$

$\text{ Area of quadrilateral }OACB=2\times \frac{1}{2}\times OB\times BC$

$\begin{array}{l}\text{ From }\mathrm{\Delta }OBC,\cot \theta =\frac{BC}{OB}=\frac{BC}{5}\\ \therefore \text{ Area of quadrilateral }OACB\end{array}$

$\begin{array}{l}=5\times 5\cot \theta \\ =25\times \frac{2\sqrt{3}}{\sqrt{13}}\\ =50\sqrt{\frac{3}{13}}\text{ sq.units }\end{array}$