z1 and z2 lie on a circle with center at the origin. The point of intersection z3 of the tangents at z1 and z2 is given by

As ΔOAC is a right-angled triangle with right angle at A, soz12+z3−z12=z32 ⇒ 2z12−z3z1−z1z3=0 ⇒ 2z1−z3−z1z1z3=0 (1)Similarly,2z2−z3−z2z2z3=0 (2)Subtracting (2) from (1), we get2z2−z1=z3z1z1−z2z2⇒ 2r2z1−z2z1z2=z3r2z22−z12z12z22 ∵z12=z22=r2⇒ z3=2z1z2z2+z1

z1 and z2 lie on a circle with center at the origin. The point of intersection z3 of the tangents at z1 and z2 is given by

As ΔOAC is a right-angled triangle with right angle at A, soz12+z3−z12=z32 ⇒ 2z12−z3z1−z1z3=0 ⇒ 2z1−z3−z1z1z3=0 (1)Similarly,2z2−z3−z2z2z3=0 (2)Subtracting (2) from (1), we get2z2−z1=z3z1z1−z2z2⇒ 2r2z1−z2z1z2=z3r2z22−z12z12z22 ∵z12=z22=r2⇒ z3=2z1z2z2+z1

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z₁ and z₂ lie on a circle with center at the origin. The point of intersection z₃ of the tangents at z₁ and z₂ is given by

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$\frac{1}{2} (z_{1} + z_{2})$

$\frac{2 z_{1} z_{2}}{z_{1} + z_{2}}$

$\frac{1}{2} (\frac{1}{z_{1}} + \frac{1}{z_{2}})$

$\frac{z_{1} + z_{2}}{z_{1} z_{2}}$

answer is B.

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Detailed Solution

As $ΔOAC$ is a right-angled triangle with right angle at A, so

${|z_{1}|}^{2} + {|z_{3} - z_{1}|}^{2} = {|z_{3}|}^{2} \Rightarrow 2 {|z_{1}|}^{2} - z_{3} z_{1} - z_{1} z_{3} = 0 \Rightarrow 2 z_{1} - z_{3} - \frac{z_{1}}{z_{1}} z_{3} = 0 (1)$

Similarly,

$2 z_{2} - z_{3} - \frac{z_{2}}{z_{2}} z_{3} = 0$ (2)

Subtracting (2) from (1), we get

$\begin{array}{l} 2 (z_{2} - z_{1}) = z_{3} (\frac{z_{1}}{z_{1}} - \frac{z_{2}}{z_{2}}) \\ \Rightarrow \frac{2 r^{2} (z_{1} - z_{2})}{z_{1} z_{2}} = z_{3} r^{2} (\frac{z_{2}^{2} - z_{1}^{2}}{z_{1}^{2} z_{2}^{2}}) [∵ {|z_{1}|}^{2} = {|z_{2}|}^{2} = r^{2}] \\ \Rightarrow z_{3} = \frac{2 z_{1} z_{2}}{z_{2} + z_{1}} \end{array}$