Question:

Step 1: Set up the constraint equation by introducing a parameter. Let $k = \dfrac{y}{x}$, which means $y = kx$. We want to find the maximum value of $k$ for points $(x, y)$ on the circle $(x-3)^2 + (y-3)^2 = 6$. Substituting $y = kx$ into the circle equation: $$(x-3)^2 + (kx-3)^2 = 6$$ Step 2: Expand and simplify the equation. Expanding both squared terms: $$x^2 - 6x + 9 + k^2x^2 - 6kx + 9 = 6$$ Combining like terms: $$(1+k^2)x^2 - 6(1+k)x + 12 = 0$$ Step 3: Apply the discriminant condition for real solutions. For the line $y = kx$ to intersect the circle at real points, the quadratic equation in $x$ must have real solutions. This requires the discriminant to be non-negative: $$\Delta = [6(1+k)]^2 - 4(1+k^2)(12) \geq 0$$ $$36(1+k)^2 - 48(1+k^2) \geq 0$$ Dividing by 12: $$3(1+k)^2 - 4(1+k^2) \geq 0$$ Expanding: $$3(1 + 2k + k^2) - 4(1 + k^2) \geq 0$$ $$3 + 6k + 3k^2 - 4 - 4k^2 \geq 0$$ $$-k^2 + 6k - 1 \geq 0$$ Multiplying by $-1$ (and reversing the inequality): $$k^2 - 6k + 1 \leq 0$$ Step 4: Solve the quadratic inequality to find the range of $k$. Using the quadratic formula to find the roots: $$k = \dfrac{6 \pm \sqrt{36-4}}{2} = \dfrac{6 \pm \sqrt{32}}{2} = \dfrac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}$$ Since the coefficient of $k^2$ is positive, the parabola opens upward, so the inequality $k^2 - 6k + 1 \leq 0$ is satisfied between the roots: $$3 - 2\sqrt{2} \leq k \leq 3 + 2\sqrt{2}$$ Step 5: Identify the maximum value. The maximum value of $k = \dfrac{y}{x}$ is: $$k_{\max} = 3 + 2\sqrt{2}$$ **Final Answer:** The largest value of $\dfrac{y}{x}$ is $\boxed{3 + 2\sqrt{2}}$, which corresponds to **Option 3**.