Circles
Optimization on Circle
Grade 11
Question:
<p>The largest value of \(\dfrac{y}{x}\), where \((x, y)\) is a real number pair satisfying \((x-3)^2 + (y-3)^2 = 6\), is:</p>
<p>(a) \(2\sqrt{3}\)</p>
<p>(b) \(2 + \sqrt{3}\)</p>
<p>(c) \(3 + 2\sqrt{2}\)</p>
<p>(d) \(6 + 2\sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: The maximum value of y/x for points on a circle equals the maximum slope of a line through the origin that is tangent to the circle. Geometrically, we need the line y = mx through origin to be tangent to the given circle.
<p><strong>Step 1:</strong> We seek the maximum value of m = y/x for points (x,y) on circle (x-3)² + (y-3)² = 6.</p><p><strong>Step 2:</strong> The line y = mx passes through origin. For maximum m, this line must be tangent to the circle.</p><p><strong>Step 3:</strong> Distance from center (3,3) to line mx - y = 0 equals radius √6:</p><p>$$\frac{|3m - 3|}{\sqrt{m^2 + 1}} = \sqrt{6}$$</p><p><strong>Step 4:</strong> Square both sides:</p><p>$$\frac{(3m-3)^2}{m^2+1} = 6$$</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 5:</strong> Using quadratic formula:</p><p>$$m = \frac{6 \pm \sqrt{36-4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}$$</p><p><strong>Step 6:</strong> The maximum value is m = 3 + 2√2</p><p>∴ Answer: B</p>
Correct Answer: B