Given $|x|= |y| = 1$, show that the intersection of two tangents to a unit circle at points $x$ and $y$ is $\frac{2xy}{x+y}$.
What would be the best place to start to attempt this problem?
What I know: $x^2= x\overline{x} = 1 = y\overline{y}$
Given $|x|=|y| = 1$, show that the intersection of two tangents to a unit circle at points $x$ and $y$ is $\frac{2xy}{x+y}$.
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geometry
complex-numbers
2 Answers
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Let $Z $ be their intersection point. Then from here, we know that $$z +x^2\bar z =2x \text { and } z+y^2\bar z =2y $$So $$(x^2-y^2)\bar z=2 (x-y)$$ $$\Rightarrow \bar z =\frac {2}{x+y} $$ Thus $z = \frac {2}{\bar x + \bar y} $. Since $x $ and $y $, lie on the unit circle, $$\bar x =\frac{1}{x} \text { and } \bar y = \frac {1}{y} $$ and the result follows. Hope it helps.
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For an alternative, more geometric proof, consider that the point of intersection $z$ must lie on the bisector of angle $\angle xoy$ by symmetry, so $z = \lambda(x+y)$ for some $\lambda \in \mathbb{R}\,$.
The power of point $z$ with respect to the unit circle is $|z|^2-1$ and, since $zx$ is a tangent, it also equals $|z-x|^2$. Therefore, using $\,\bar x = 1/x\,$ and $\,\bar y = 1/y\,$ at the last step, since $|x|=|y|=1\,$:
$$ \require{cancel} \begin{align} z \bar z -1 & = (z-x)(\bar z - \bar x) \\ \cancel{z \bar z} -1 & = \cancel{z \bar z} - z \bar x - \bar z x + x \bar x \\ z \bar x + \bar z x & = 2 \\ \lambda(x+y)\bar x + \lambda (\bar x + \bar y) x & = 2 \end{align} $$ $$ \lambda = \frac{2}{x \bar x + \bar x y+\bar x x +x \bar y} = \frac{2}{2 + \frac{x}{y}+\frac{y}{x}} = \frac{2xy}{x^2+y^2+2xy} = \frac{2xy}{(x+y)^2} $$
Then $z = \lambda (x+y)=\cfrac{2xy}{x+y}\,$.