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If the chord of contact of the point $\mathrm{P}(1,1)$ with respect to the circle $S=x^2+y^2+4 x+6 y-3=0$ meet the circle $\mathrm{S}=0$ at $\mathrm{A}$ and $\mathrm{B}$, then the area of $\triangle \mathrm{PAB}$ is
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$\frac{108}{25}$
$\begin{aligned} & \text { } S=x^2+y^2+4 x+6 y-3=0 \\ & \Rightarrow(x+2)^2+(y+2)^2=4^2\end{aligned}$
$\begin{aligned} & \mathrm{OP}^2=\mathrm{AP}^2+\mathrm{OA}^2 \Rightarrow 5^2=4^2+\mathrm{AP}^2 \Rightarrow \mathrm{AP}=3 \\ & \text { Let } \angle \mathrm{AOP}=\angle \mathrm{BOP}=\theta \\ & \sin \theta=\frac{3}{5} \Rightarrow \cos \theta=\frac{4}{5} \\ & \sin 2 \theta=2 \sin \theta \cos \theta=2 \times \frac{3}{5} \times \frac{4}{5}=\frac{24}{25} \\ & \Delta \mathrm{AOB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB} \sin 2 \theta \\ & =\frac{1}{2} \times 4 \times 4 \times \frac{24}{25}=\frac{192}{25} \\ & \Rightarrow \Delta \mathrm{OAP}=\Delta \mathrm{AOP}-\Delta \mathrm{AOQ} \\ & =\Delta \mathrm{AOP}-\frac{\Delta \mathrm{AOB}}{2} \\ & =\frac{1}{2} \times 4 \times 3-\frac{96}{25}=6-\frac{96}{25}=\frac{54}{25} \\ & \Rightarrow \Delta \mathrm{PAB}=2 \times \Delta \mathrm{OAP}=2 \times \frac{54}{25}=\frac{108}{25}\end{aligned}$
$\begin{aligned} & \mathrm{OP}^2=\mathrm{AP}^2+\mathrm{OA}^2 \Rightarrow 5^2=4^2+\mathrm{AP}^2 \Rightarrow \mathrm{AP}=3 \\ & \text { Let } \angle \mathrm{AOP}=\angle \mathrm{BOP}=\theta \\ & \sin \theta=\frac{3}{5} \Rightarrow \cos \theta=\frac{4}{5} \\ & \sin 2 \theta=2 \sin \theta \cos \theta=2 \times \frac{3}{5} \times \frac{4}{5}=\frac{24}{25} \\ & \Delta \mathrm{AOB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB} \sin 2 \theta \\ & =\frac{1}{2} \times 4 \times 4 \times \frac{24}{25}=\frac{192}{25} \\ & \Rightarrow \Delta \mathrm{OAP}=\Delta \mathrm{AOP}-\Delta \mathrm{AOQ} \\ & =\Delta \mathrm{AOP}-\frac{\Delta \mathrm{AOB}}{2} \\ & =\frac{1}{2} \times 4 \times 3-\frac{96}{25}=6-\frac{96}{25}=\frac{54}{25} \\ & \Rightarrow \Delta \mathrm{PAB}=2 \times \Delta \mathrm{OAP}=2 \times \frac{54}{25}=\frac{108}{25}\end{aligned}$
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