Find the largest value of $y/x$ for a point $(x,y)$ on the circle $(x-3)^2+(y-3)^2=6$.
Step-by-Step Solution
Key Concept: Maximize slope $m=y/x$ of line through origin tangent to circle; distance from $(3,3)$ to line $y=mx$ equals $\sqrt{6}$.
$\frac{|3m-3|}{\sqrt{m^2+1}}=\sqrt{6}$. $(3m-3)^2=6(m^2+1)$. $9m^2-18m+9=6m^2+6$. $3m^2-18m+3=0$. $m^2-6m+1=0$. $m=3\pm 2\sqrt{2}$. Maximum value $=3+2\sqrt{2}$.
Correct Answer: C