# Finding point of intersection of tangents through complex numbers

If $P$ and $Q$ be two points on the circle $|w|=r$ represented by $w_1$ and $w_2$ respectively, then find the complex number representing the point of intersection of the tangents at $P$ and $Q$.

My attempt: Well frankly i didnt even know how to start solving this problem. I tried converting the problem into vectors and I found the vector equations of the tangents and tried solving them but my answer was nowhere close to the one given in my book.

Could anyone please suggest a method on how to proceed with this question.

Thanks.

• Tell us the answer in your book & show us your vector solution ... hopefully we can get them to agree. – Donald Splutterwit Sep 23 '17 at 14:21
• Well actually the book has only 4 options given(no solution) and all of them have some combination of the complex numbers w1 and w2 in the denominator of the expression. But by my vector method its impossible to arrive at such an expression. – physics123 Sep 23 '17 at 14:33

## 2 Answers

Let $p$ and $q$ be the complex numbers representing $P$ and $Q$. Since $P$ and $Q$ lie on the circle $\lvert w\rvert = r$, it holds $\lvert p\rvert = \lvert q\rvert = r$, which can be written as $p\bar{p} = q\bar{q} = r^2$ (since $z\bar{z} = \lvert z\rvert^2$ for every complex number $z$).

Now we want to find the equation of the tangent at $P$. Let $X$ be an arbitrary point in the plane and $x$ the complex number representing $X$. Also, let $O$ be the origin. $X$ belongs to the tangent iff $OP \perp PX$, that is to say, iff \begin{align*} & \Re\left(\frac{x - p}{p - 0}\right) = 0 \\ \iff & \frac{x - p}{p} = -\overline{\frac{x - p}{p}} = -\frac{\bar{x} - \bar{p}}{\bar{p}} = -\frac{\bar{x} - \frac{r^2}{p}}{\frac{r^2}{p}} = \frac{r^2 - p\bar{x}}{r^2} \\ \iff & r^2x + p^2\bar{x} - 2pr^2 = 0 \tag{1} \end{align*} So $(1)$ is the equation we were looking for.

In a similar way we get that the equation of the tangent at $Q$ is $$r^2x + q^2\bar{x} - 2qr^2 = 0 \tag{2}$$

Now, in order to determine the intersection of the two tangents, we just need to solve for $x$ (and $\bar{x}$) the system given by $(1)$ and $(2)$. Subtractiong $(2)$ from $(1)$ gives $(p^2 - q^2)\bar{x} - 2r^2(p - q) = 0$ which, cancelling a factor $p - q \ne 0$, becomes $(p + q)\bar{x} - 2r^2 = 0$, whence $\displaystyle \bar{x} = \frac{2r^2}{p + q}$. Subtitution in $(1)$ finally gives $$x = 2p - \frac{p^2\bar{x}}{r^2} = 2p - \frac{2p^2}{p + q} = \boxed{\frac{2pq}{p + q}}$$

Edit: Ok, I just read that actually $p$ and $q$ were already defined as $w_1$ and $w_2$...

Given points $A(z_1)$ and $B(z_2)$ in plane, all points $z$ on line $AB$ have this property that $\arg(z-z_1)=\arg(z_2-z_1)$ then the equation of the line $AB$ is $$\arg\dfrac{z-z_1}{z_2-z_1}=0\tag{1}$$ Let $MP$ and $MQ$ are lines tangent to the circle $C$ with center at origin. From geometry we know $M(z)$ the point of intersection of the tangents at $P(w_1)$ and $Q(w_2)$ makes bisector $MO$ and perpendicular to $PQ$. With vectors $\vec{OP}+\vec{OQ}=\vec{OM}$ or $w_1+w_2=z$. Thus from (1), the locus of all $M$ (passing through origin $O(0)$) is $$\arg\dfrac{z-0}{w_1+w_2-0}=0$$ or $$\arg\dfrac{z}{w_1+w_2}=0$$