Let coordinate of point $\mathrm{A}\left(\mathrm{t}^2, 2 \mathrm{t}\right)(\square \mathrm{a}=1)$
equation of tangent at point A $$ \begin{aligned} & y t=x+t^2 \\ & x-t y+t^2=0 \end{aligned} $$ centre of circle $(3,0)$
Now PD = radius $$ \begin{aligned} & \left|\frac{3-0+t^2}{\sqrt{1+t^2}}\right|=3 \\ & \left(3+t^2\right)^2=9\left(1+t^2\right) \\ & 9+t^4+6 t^2=9+9 t^2 \\ & t=0,-\sqrt{3}, \sqrt{3} \end{aligned} $$ So point $$ \mathrm{A}(3,2 \sqrt{3}) $$ $$ \Rightarrow \quad a=3, b=2 \sqrt{3} $$ (Since it lies in first quadrant)
For point B which is foot of perpendicular from centre $(3,0)$ to the tangent $x-\sqrt{3} y+3=0$ $$ \begin{aligned} & \frac{c-3}{1}=\frac{d-0}{-\sqrt{3}}=\frac{-(3-0+3)}{4} \\ & \Rightarrow c=\frac{3}{2} \quad d=\frac{3 \sqrt{3}}{2} \\ & \Rightarrow 2\left(\frac{3}{2}+3\right)=9 \end{aligned} $$