<s>[INST] Find the radius of the circle given in the picture below. This is the image of the question. I am not able to get how to find the radius. Please help with that.

This is my try. I can't proceed now after it.

Thanks
 [/INST] 
$$PT^2=PC\cdot PA\implies PT^2=2\cdot 5=10$$
Using Pythagorean theorem in right $\Delta PTO$,   $PO=\sqrt{OT^2+PT^2}=\sqrt{R^2+10}$
Using cosine rule in $\Delta ABP$,   $$\cos\angle APB=\frac{PB^2+PA^2-AB^2}{2(PB)(PA)}=\frac{(\sqrt{R^2+10}+R)^2+5^2-(\sqrt{11})^2}{2(\sqrt{R^2+10}+R)(5)}\tag1$$
Similarly, using cosine rule in $\Delta AOP$,   $$\cos\angle APB=\frac{PO^2+PA^2-OA^2}{2(PO)(PA)}=\frac{(\sqrt{R^2+10})^2+5^2-(R)^2}{2(\sqrt{R^2+10})(5)}\tag2$$
Equating the values of $\cos \angle APB$ from (1) & (2),
$$\frac{(\sqrt{R^2+10}+R)^2+5^2-(\sqrt{11})^2}{2(\sqrt{R^2+10}+R)(5)}=\frac{(\sqrt{R^2+10})^2+5^2-(R)^2}{2(\sqrt{R^2+10})(5)}$$
$$\frac{2R\sqrt{R^2+10}+2R^2+24}{\sqrt{R^2+10}+R}=\frac{35}{\sqrt{R^2+10}}$$
$$(11-2R^2)\sqrt{R^2+10}=2R^3-15R$$
$$(11-2R^2)^2(\sqrt{R^2+10})^2=(2R^3-15R)^2$$
$$28R^4-272R^2+605=0$$
$$R^2=\frac{-(-272)\pm\sqrt{(-272)^2-4\cdot 28\cdot605}}{2\cdot 56}=\frac{68\pm\sqrt{389}}{14}$$
$$R=\sqrt{\frac{68+\sqrt{389}}{14}}\approx 2.50318487$$
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