$P$ is the moving point on the circle $O: x^2+y^2=1$, the tangent line of circle $O$ through $P$ intersects with circle $O_2:(x-1)^2+(y-4)^2=6^2$ two points $A$, $B$. What is the maximum value of $\frac{|PA|}{|PB|}$?
I have tried setting coordinates of $P$ is $(cos(\theta),sin(\theta))$, and the equation of line $AB$ is $x \sin (\theta )+y \cos (\theta )-1=0$, simultaneous equations of line $AB$ and circle $O_2$ to find the coordinates of $A$,$B$, this is too complicated, is there a better way?
Let $Q_2K$ be a perpendicular to $AB$ and let $O_2K=x$.
Thus, since $OO_2=\sqrt{17}$ and $OP=1,$ we obtain: $$\frac{PA}{PB}=\frac{AK+PK}{AK-PK}=\frac{\sqrt{36-x^2}+\sqrt{17-(x-1)^2}}{\sqrt{36-x^2}-\sqrt{17-(x-1)^2}}=$$ $$=\frac{52+2x-2x^2+2\sqrt{(36-x^2)(16+2x-x^2)}}{20-2x}=$$ $$=\frac{52+2x-2x^2+\sqrt2\sqrt{(36-x^2)(32+4x-2x^2)}}{20-2x}\leq$$ $$\leq\frac{52+2x-2x^2+\frac{1}{\sqrt2}(36-x^2+32+4x-2x^2)}{20-2x}\leq3+2\sqrt2,$$ where the last inequality it's $$(x-2)^2\geq0.$$
The equality occurs for $x=2$, which says that $3+2\sqrt2$ is a maximal value.