Find the largest value for $\frac{y}{x}$ for pairs of [real numbers](https://artofproblemsolving.com/wiki/index.php/Real_numbers) $(x, y)$ which satisfy $(x-3)^2+(y-3)^2=6$.
$\mathrm{(A) \ }3+2\sqrt{2} \qquad \mathrm{(B) \ }2+\sqrt{3} \qquad \mathrm{(C) \ } 3\sqrt{3} \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 6+2\sqrt{3}$
Let $\frac{y}{x}=k$, so that $y=kx$. Substituting this into the given [equation](https://artofproblemsolving.com/wiki/index.php/Equation) $(x-3)^2+(y-3)^2=6$ yields $(x-3)^2+(kx-3)^2=6$. Multiplying this out and forming it into a [quadratic](https://artofproblemsolving.com/wiki/index.php/Quadratic) yields $(1+k^2)x^2+(-6-6k)x+12=0$. 
We want $x$ to be a [real number](https://artofproblemsolving.com/wiki/index.php/Real_number), so we must have the [discriminant](https://artofproblemsolving.com/wiki/index.php/Discriminant) $\geq0$. 
The discriminant is $(-6-6k)^2-4(1+k^2)(12)=36k^2+72k+36-48-48k^2=-12k^2+72k-12$. Therefore, we must have $-12k^2+72k-12\geq0$, or $k^2-6k+1\leq0$. The [roots](https://artofproblemsolving.com/wiki/index.php/Root_(polynomial)) of this quadratic, using the [quadratic formula](https://artofproblemsolving.com/wiki/index.php/Quadratic_formula), are $3\pm2\sqrt{2}$, so the quadratic can be [factored](https://artofproblemsolving.com/wiki/index.php/Factoring) as $(k-(3-2\sqrt{2}))(k-(3+2\sqrt{2}))\leq0$. We can now separate this into $3$ cases:
Case 1: $k3+2\sqrt{2}$
Then, both terms are positive, so the inequality doesn't hold.
Also, when $k=3-2\sqrt{2}$ or $k=3+2\sqrt{2}$, the equality holds.
Therefore, we must have $3-2\sqrt{2}\leq k\leq3+2\sqrt{2}$, and the maximum value of $k=\frac{y}{x}$ is ${3+2\sqrt{2}}, \boxed{\text{A}}$.