Complex Numbers
Tangent to Circle in Complex Plane
Grade 11
Question:
<p>Equation of tangent drawn to circle \(|z| = r\) at the point \(A(z_0)\) is</p>
<p>(1) \(\text{Re}\left(\dfrac{z}{z_0}\right) = 1\)</p>
<p>(2) \(z\bar{z}_0 + z_0\bar{z} = 2r^2\)</p>
<p>(3) \(\text{Im}\left(\dfrac{z}{z_0}\right) = 1\)</p>
<p>(4) \(\text{Im}\left(\dfrac{z_0}{z}\right) = 1\)</p>
Step-by-Step Solution
Key Concept: The tangent to a circle |z| = r at point z₀ is perpendicular to the radius OA, so the tangent line satisfies Re(z·z̄₀) = |z₀|² = r². This comes from the condition that (z - z₀) ⊥ z₀.
<p><strong>Step 1:</strong> For circle |z| = r, the point A is at z₀ where |z₀| = r.</p><p><strong>Step 2:</strong> The tangent at z₀ is perpendicular to the radius vector z₀. Any point z on the tangent satisfies: (z - z₀) ⊥ z₀, meaning Re[(z - z₀)·z̄₀] = 0.</p><p><strong>Step 3:</strong> Expanding: Re[z·z̄₀ - z₀·z̄₀] = 0 → Re[z·z̄₀] = Re[|z₀|²] = r² (since |z₀| = r).</p><p><strong>Step 4:</strong> Alternative form: Re[z·z̄₀] = r² or z·z̄₀ + z̄·z₀ = 2r² (Cartesian form).</p><p>∴ Answer: <strong>Re(z·z̄₀) = r²</strong> or equivalently <strong>z·z̄₀ + z̄·z₀ = 2r²</strong></p>
Correct Answer: 1