Question:

<p><strong>Step 1:</strong> Let y/x = m, so y = mx. We need the line y = mx through the origin to be tangent to the circle (x-3)² + (y-3)² = 6.</p><p><strong>Step 2:</strong> Rewrite as mx - y = 0. The distance from center (3,3) to this line must equal the radius √6:</p><p>$$\frac{|3m - 3|}{\sqrt{m^2 + 1}} = \sqrt{6}$$</p><p><strong>Step 3:</strong> Square both sides: $$\frac{(3m-3)^2}{m^2+1} = 6$$</p><p>$$(3m-3)^2 = 6(m^2+1)$$</p><p>$$9m^2 - 18m + 9 = 6m^2 + 6$$</p><p>$$3m^2 - 18m + 3 = 0$$</p><p>$$m^2 - 6m + 1 = 0$$</p><p><strong>Step 4:</strong> Using the quadratic formula: $$m = \frac{6 ± \sqrt{36-4}}{2} = \frac{6 ± \sqrt{32}}{2} = \frac{6 ± 4\sqrt{2}}{2} = 3 ± 2\sqrt{2}$$</p><p><strong>Step 5:</strong> The larger value is m = 3 + 2√2, which represents the maximum slope (and thus maximum y/x).</p><p>∴ Answer: <strong>3 + 2√2</strong></p>