Multiple option question based on tangents to a circle

Question

Let tangents at $A(z_{1})$ and $B(z_{2})$ be drawn to the circle $|z|=2$. Then which of the following is/are correct?

$(a)\;$ Equation of tangents at $A(z_{1})$ is given by $z+\bar{z}=2z_{1}$.

$(b)\; $ If the tangents at $A(z_{1})$ and $B(z_{2})$ Intersect at $z_{3}$, then $\displaystyle z_{3}=\frac{2z_{1}z_{2}}{z_{1}+z_{2}}$.

$(c)\; $ Slope of tangent at$A(z_{1})$ is $\displaystyle \frac{1}{i}\bigg(\frac{z+\bar{z_{1}}}{z_{1}-\bar{z_{1}}}\bigg)$.

$(d)\;$ If the points $A(z_{1})$ and $B(z_{2})$ on the circle $|z|=2$ such that $z_{1}+z_{2}=0$, then tangents intersects at $90^\circ$.

For solution I tried to assume $O$ is the center of the circle $|z|=2$ and $P(z_{3})$ is a point outside the circle. Rotation theorem for $\triangle POA$ gives

$$(z_{1}-z_{3})=-z_{3}e^{i\theta}, \tag{1}$$

and rotation theorem for $\triangle POB$ gives

$$(z_{2}-z_{3})=-z_{3}e^{-i\theta}. \tag{2}$$

From these two equations, $(z_{1}-z_{3})(z_{2}-z_{3})=z^2_{3}$, so

$$z_{3}=\frac{z_{1}z_{2}}{z_{1}+z_{2}}.$$

If $z_{1}+z_{2}=0$ . Then $z_{3}=\infty$ . so $\arg(z_{3})=90^\circ$.

Please help me to find if other options are right or wrong.

Answer 1

$\def\i{\mathrm{i}}$Option (a): For any $P(z)$ to be on the tangent line at $A(z_1)$, it is equivalent to satisfy $PA ⊥ OA$, or $\dfrac{z - z_1}{z_1} = \i k$ for some $k \in \mathbb{R}$, or$$ \frac{z - z_1}{z_1} + \frac{\overline{z} - \overline{z_1}}{\overline{z_1}} = 0, $$ or (note that $|z_1| = 2$)$$ \overline{z_1} z + z_1 \overline{z} = 2 z_1 \overline{z_1} = 8. $$

Option (b): Using the equation in (a),$$ \left\{\begin{array}{l} \overline{z_1} z_3 + z_1 \overline{z_3} = 8\\ \overline{z_2} z_3 + z_2 \overline{z_3} = 8 \end{array}\right. $$ This is a linear system of equations with respect to $z_3$ and $\overline{z_3}$, and solving for $z_3$ to get$$ z_3 = \frac{8 z_2 - 8 z_1}{\overline{z_1} z_2 - \overline{z_2} z_1} = \frac{8 (z_2 - z_1)}{\dfrac{4 z_2}{z_1} - \dfrac{4 z_1}{z_2}} = 2 z_1 z_2 · \frac{z_2 - z_1}{z_2^2 - z_1^2} = \frac{2 z_1 z_2}{z_1 + z_2}. $$

Option (c): suppose $z_1 = x + \i y$, where $x, y \in \mathbb{R}$.
If $x = 0$, then $y = \pm 2$ and the slope of the tangent line at $A(z_1)$ is $0$.
If $y = 0$, then the slope of the tangent line at $A(z_1)$ does not exist.
If $x, y \neq 0$, then the slope of $OA$ is $\dfrac{y}{x}$, which implies the slope at the tangent line at $A(z_1)$ is $-\dfrac{x}{y}$ and$$ -\frac{x}{y} = -\frac{\dfrac{1}{2} (z_1 + \overline{z_1})}{\dfrac{1}{2\i} (z_1 - \overline{z_1})} = \frac{1}{\i} \frac{z_1 + \overline{z_1}}{z_1 - \overline{z_1}}. $$ Therefore, if $z \not\in \mathbb{R}$, then the slope of the tangent line at $A(z_1)$ is $\dfrac{1}{\i} \dfrac{z_1 + \overline{z_1}}{z_1 - \overline{z_1}}$.

Option (d): If $z_1 + z_2 = 0$, from Option (b) it can be known that the tangent line at $A(z_1)$ and that at $B(z_2)$ does not intersect, i.e. they are parallel.

Answer 2

Option (a) is obviously incorrect as $z_1$ does not satisfy $z+\bar z=2z_1$.

$A(z_1)$ is the mid-point of $2z_1$ and $0$. The tangent at $A(z_1)$ to the circle is the perpendicular bisector of $2z_1$ and $0$. So its equation is

\begin{align*} |z-0|&=|z-2z_1|\\ z\bar z&=(z-2z_1)(\bar z-2\bar z_1)\\ z\bar z&=z\bar z-2z\bar z_1-2\bar zz_1+4z_1\bar z_1\\ z\bar z_1+\bar zz_1&=2z_1\bar z_1\\ \frac{z}{z_1}+\frac{\bar z}{\bar z_1}&=2 \end{align*}

The real part of $\displaystyle \frac{z}{z_1}$ is $1$.

The equation of the tangent at $B(z_2)$ is $\displaystyle \frac{z}{z_2}+\frac{\bar z}{\bar z_2}=2$.

$z_3$ can be found by solving the equations of the two tangents. Alternatively, we can check whether $\displaystyle z=\frac{2z_1z_2}{z_1+z_2}$ satisfies the equation $\displaystyle \frac{z}{z_1}+\frac{\bar z}{\bar z_1}=2$. Indeed

\begin{align*} \frac{2z_2}{z_1+z_2}+\frac{2\bar z_2}{\bar z_1+\bar z_2}&=\frac{2(z_2\bar z_1+z_2\bar z_2+z_1\bar z_2+z_2\bar z_2)}{z_1\bar z_1+z_1\bar z_2+z_2\bar z_1+z_2\bar z_2}\\ &=\frac{2(z_2\bar z_1+4+z_1\bar z_2+4)}{4+z_1\bar z_2+z_2\bar z_1+4}\\ &=2 \end{align*}

So, $\displaystyle z=\frac{2z_1z_2}{z_1+z_2}$ satisfies the equation of the tangent at $A(z_1)$. Similarly it also satisfies the equation of the tangent at $B(z_2)$.

Option (b) is correct.

Option (c) looks strange, as the slope should not be a function of $z$. I think there is a typo and I suppose that it is $\displaystyle \frac{1}{i}\bigg(\frac{z_1+\bar{z_{1}}}{z_{1}-\bar{z_{1}}}\bigg)$.

The tangent at $A(z_1)$ is the perpendicular bisector of $0$ and $2z_1$. So, it is in a direction perpendicular to $2z_1-0=2z_1$. So it is in the direction of $iz_1$. Its slope is

$$\frac{\Im(iz_1)}{\Re(iz_1)}=\frac{\frac{1}{2i}(iz_1-\bar i\bar z_1)}{\frac{1}{2}(iz_1+\bar i\bar z_1)}= \frac{1}{i}\bigg(\frac{z_1+\bar{z_{1}}}{z_{1}-\bar{z_{1}}}\bigg)$$

However, when $z_1$ is real, the tangent is parallel to the imaginary axis and has no slope.

So, option (c) is correct when $z_1\ne 2$ and $z_1\ne -2$.

Option (d) is incorrect. The tangent at $A(z_1)$ to the circle is the perpendicular bisector of $2z_1$ and $0$, while the tangent at $B(z_2)$ to the circle is the perpendicular bisector of $2z_2$ and $0$. $z_1+z_2=0$ implies that $2z_1$, $0$ and $2z_2$ are collinear. So the two tangents are parallel.