analytic geometrycirclesgeometrytangent line
I have the following problem: there is a circle with the $R = 5$ and center of the circle located on coordinate $(0, 0)$. I have two points $A(6, 8)$ and $B(-4, -6)$. From points, tangents to the circle were drawn. It is better illustrated as:
Let us denote points where tangents and circle intersect as $E, F, G, H$ (see the picture above for better the understandings). So we need to find the distance between E and F along the circle.
Yes. Let $O$ be the centre of the circle, $P$ and $Q$ the points on the circumference, and $M$ the midpoint of $\overline{PQ}$; then $\triangle POM$ is a right triangle. Let $\theta=\angle POM$, and let $r=|OP|$, the radius of the circle. Then $\sin\theta=\dfrac{|PM|}r$. Since $\theta$ is half the known angle, and $|PM|$ is half the known distance, both $\sin\theta$ and $|PM|$ are also known, and we can calculate $r$.
Specifically, if $\alpha$ is the known angle, and $d$ is the known distance, then
$$r=\frac{d/2}{\sin(\alpha/2)}=\frac{d}{2\sin(\alpha/2)}\;.$$
Here is a picture:
$\overline {CP} = d$
$\angle CAP$ and $\angle BAP$ are right angles, and $\triangle APB$ is isosceles.
$m\angle APC = \arcsin \frac rd\\ m\angle APB = 2\arcsin \frac rd\\ m\angle BAP = \arccos \frac rd$
Best Answer
\begin{align} |OE|=|OF|= R&=5 ,\quad |OA|=10 ,\quad |OB|=2\sqrt{13} ,\quad |AB|=2\sqrt{74} ,\\ \triangle AOE:\quad |AE|&=5\sqrt3 ,\\ \triangle BFO:\quad |BF|&=3\sqrt3 . \end{align}
\begin{align} \angle EOF&=\angle AOB-\angle AOE-\angle FOB , \end{align}
\begin{align} \angle AOB&=\arccos\frac{|OA|^2+|OB|^2-|AB|^2}{2\cdot|OA|\cdot|OB|} = \pi-\arccos(\tfrac{18}{65}\sqrt{13}) ,\\ \angle AOE&= \arccos\frac{|OE|}{|OA|} =\tfrac\pi3 ,\\ \angle FOB&= \arccos\frac{|OF|}{|OB|} =\arccos(\tfrac5{26}\sqrt{13}) ,\\ \angle EOF&= \tfrac{2\pi}3-\arccos(\tfrac{18}{65}\sqrt{13}) -\arccos(\tfrac5{26}\sqrt{13}) \approx 1.234262917 . \end{align}
So, the distance between $E$ and $F$ along the circle, that is, the length of the arc $FE$ is
\begin{align} R\cdot\angle EOF&= 5\cdot(\tfrac{2\pi}3-\arccos(\tfrac{18}{65}\sqrt{13}) -\arccos(\tfrac5{26}\sqrt{13})) \approx 6.171314600 . \end{align}
Expression for $\angle EOF$ can be simplified to \begin{align} \angle EOF&= \arccos\frac{18+2\sqrt3}{65} , \end{align} hence by the cosine rule we can also find
\begin{align} |EF|&=\tfrac1{13}\sqrt{6110-260\sqrt3} \approx 5.78698130 . \end{align}