Point of intersection of the tangents at two points on a circle

Question

The point of intersection of the tangents at the points $P(\theta_1)$ and $Q(\theta_2)$ on the circle $x^2+y^2=a^2$ are $$ x=\frac{a\cos(\frac{\theta_1+\theta_2}{2})}{\cos(\frac{\theta_1-\theta_2}{2})}\;;\;y=\frac{a\sin(\frac{\theta_1+\theta_2}{2})}{\cos(\frac{\theta_1-\theta_2}{2})}\\ $$

$$ \vec{PQ}\perp \vec{OC}\implies \bigg<a(\cos\theta_1-\cos\theta_2),r(\sin\theta_1-\sin\theta_2)\bigg>\cdot<x,y>=0\\ -2ax\sin(\frac{\theta_1+\theta_2}{2}\sin(\frac{\theta_1-\theta_2}{2}))+2ay\cos(\frac{\theta_1+\theta_2}{2}\sin(\frac{\theta_1-\theta_2}{2}))=0 $$ but it is not leading to anywhere and I really don't see any hint on how to approach the problem ?

Answer 1

The slopes of OP and OQ are $\tan\theta_1$ and $\tan\theta_2$, respectively. Given that $P=a(\cos\theta_1,\sin\theta_1)$ and $Q=a(\cos\theta_2,\sin\theta_2)$, the equations for the two tangent lines PC and QC are,

$$ (y-a\sin\theta_1) = -\frac{1}{\tan\theta_1}(x-a\cos\theta_1)$$ $$ (y-a\sin\theta_2) = -\frac{1}{\tan\theta_2}(x-a\cos\theta_2)$$

After simplification,

$$ x\cos\theta_1 + y \sin\theta_1 =a$$ $$ x\cos\theta_2 + y \sin\theta_2 =a$$

Then, just solve for the intersection to obtain,

$$x= a\frac{\sin\theta_2-\sin\theta_1}{\sin(\theta_2-\theta_1)}=a\frac{\cos\frac{\theta_1+\theta_2}{2}}{\cos\frac{\theta_1-\theta_2}{2}}$$ $$y= a\frac{\cos\theta_2-\cos\theta_1}{\sin(\theta_2-\theta_1)}=a\frac{\sin\frac{\theta_1+\theta_2}{2}}{\cos\frac{\theta_1-\theta_2}{2}}$$

Answer 2

Your last equation leads to being able to deduce the ratio $\frac yx.$ That is about as much as you could expect from it, since the condition $\vec{PQ}\perp \vec{OC}$ tells you only about the direction of $OC,$ not the length.

What about $\vec{OP}\perp \vec{PC}$ and $\vec{OQ}\perp \vec{QC}$?

Answer 3

Hint

How about first rotating and scaling the system so that $P$ and $Q$ are symmetrical - as in your drawing - with respect to the $x$ axis and on the unit circle?

In this new reference $P$ has coordinates

$$P\left(\cos\frac{\theta_1-\theta_2}2,\sin\frac{\theta_1-\theta_2}2\right)$$

and $Q$ has coordinates

$$P\left(\cos\frac{\theta_1-\theta_2}2,-\sin\frac{\theta_1-\theta_2}2\right).$$

Let $PH$ be the altitude of $\triangle OPC$ with respect to $OC$. By Euclid's first Theorem on $\triangle OPC$,

$$\overline{OC} = \frac{\overline{OP}^2}{\overline{OH}}=\frac1{\overline{OH}},$$

meaning that $C$ has coordinates $$C\left(\frac1{\cos\frac{\theta_1-\theta_2}2},0\right).$$

Now you can scale by the factor $a$ and rotate by the angle $\frac{\theta_1+\theta_2}2$ to get your result.

Answer 4

The relations are seen directly in the diagram with careful observation.

Diagonal of kite OC bisects $\angle POQ$

Let half difference of angles be $$ \angle POC=\delta =\frac{\theta_1 - \theta_2}{2}$$ The hypotenuse $H$ in right triangle $OPC$ is $$H=a \sec \delta$$

Let half sum or average angle made by angle bisector to $x-$ axis be $$\nu =\frac{\theta_1 + \theta_2}{2}$$

The $x,y$ projection on $(x,y)$ axes are $$ x= H \cos \nu,\, y = H \sin \nu. $$