<p>\(z_1\) and \(z_2\) lie on a circle with center at the origin. The point of intersection \(z_3\) of the tangents at \(z_1\) and \(z_2\) is given by</p>
<p>\(\frac{1}{2}(\bar{z}_1 + \bar{z}_2)\)</p>
<p>\(\frac{2z_1 z_2}{z_1 + z_2}\)</p>
<p>\(\frac{1}{2}\left(\frac{1}{\bar{z}_1} + \frac{1}{\bar{z}_2}\right)\)</p>
<p>\(\frac{z_1 + z_2}{\bar{z}_1 \bar{z}_2}\)</p>
Step-by-Step Solution
Key Concept: For a circle centered at origin with radius r, the tangent at point z₁ is perpendicular to the radius Oz₁. If two tangents from external point z₃ touch the circle at z₁ and z₂, then z₃ lies on the polar line, and the relationship z₃ = r²/(z₁*z₂*) holds when |z₁| = |z₂| = r.
<p><strong>Step 1:</strong> Since z₁ and z₂ lie on a circle with center O (origin) and radius r, we have |z₁| = |z₂| = r.</p><p><strong>Step 2:</strong> The tangent at z₁ is perpendicular to radius Oz₁. Tangent at z₁ has direction perpendicular to z₁, so any point on this tangent can be written as z₁ + t·iz₁ for real parameter t.</p><p><strong>Step 3:</strong> Similarly, tangent at z₂ is z₂ + s·iz₂ for real parameter s.</p><p><strong>Step 4:</strong> At intersection point z₃, we require: the line from z₁ to z₃ is perpendicular to Oz₁, and line from z₂ to z₃ is perpendicular to Oz₂.</p><p><strong>Step 5:</strong> This means (z₃ - z₁) ⊥ z₁ and (z₃ - z₂) ⊥ z₂, giving us Re(z₃·z̄₁) = |z₁|² = r² and Re(z₃·z̄₂) = |z₂|² = r².</p><p><strong>Step 6:</strong> From the polar relationship of a circle, the locus of intersection points of tangents at two points on the circle with |z₁| = |z₂| = r is given by:</p><p><strong>z₃ = r²/(z̄₁·z̄₂)·(z₁ + z₂)</strong> or equivalently <strong>z₃ = r²·z̄₁z̄₂/(|z₁|²|z₂|²) = r²(z̄₁ + z̄₂)/(r⁴)</strong></p><p>The standard result: <strong>z₃ = (z₁z₂)/(z̄₁ + z̄₂)·r²</strong> or <strong>1/z₃ = (1/z̄₁ + 1/z̄₂)·(1/r²)</strong></p><p>∴ Answer: B</p>
Correct Answer: B