<p>Let \(x\), \(y\), \(z\) and \(t\) be real numbers such that \((x, y)\) lies on a circle having radius 3; \((z, t)\) lies on a circle having radius 2 and \(xt - yz = 6\). Find the greatest value of \(P = xz\).<br>[Note: Both circles have centre at origin.]</p>
Step-by-Step Solution
Key Concept: Use parametric representation for points on circles: (x,y) = (3cosα, 3sinα) and (z,t) = (2cosβ, 2sinβ), then apply the constraint xt - yz = 6 to find a relationship between α and β, finally optimize xz = 6cosαcosβ.
<p><strong>Step 1:</strong> Parametrize the circles. Since (x,y) lies on circle x² + y² = 9 and (z,t) lies on circle z² + t² = 4, write:</p><p>x = 3cosα, y = 3sinα</p><p>z = 2cosβ, t = 2sinβ</p><p><strong>Step 2:</strong> Apply the constraint xt - yz = 6:</p><p>3cosα · 2sinβ - 3sinα · 2cosβ = 6</p><p>6(cosαsinβ - sinαcosβ) = 6</p><p>sin(β - α) = 1</p><p>∴ β - α = π/2, so β = α + π/2</p><p><strong>Step 3:</strong> Express P = xz in terms of α:</p><p>P = 3cosα · 2cos(α + π/2)</p><p>P = 6cosα · (-sinα)</p><p>P = -6cosαsinα = -3sin(2α)</p><p><strong>Step 4:</strong> Maximize P = -3sin(2α):</p><p>The maximum value of -3sin(2α) occurs when sin(2α) = -1</p><p>Maximum P = -3(-1) = 3</p><p>∴ <strong>Answer: 3</strong></p>
Correct Answer: 3