Prove that a string cant be outside a circle How can I prove that a chord can't be outside the circle itself.
Is there a way to prove that you can't draw a chord outside the circle.
 A: Let $P_1 = (x_1,y_1)$ and $P_2 = (x_2,y_2)$ be points on the unit circle, and let $P=(x,y)$ be a point on the chord $P_1P_2$, not equal to $P_1$ or $P_2$. So we can write $P=tP_1 + (1-t)P_2$ for some $t \in (0,1)$.
So we have $x = tx_1 + (1-t)x_2$ and $y = ty_1 + (1-t)y_2$, giving
$$\begin{align}
x^2+y^2 & = t^2(x_1^2+y_1^2) + (1-t)^2(x_2^2+y_2^2) + 2t(1-t)(x_1x_2+y_1y_2) \\
& = t^2+(1-t)^2 + 2t(1-t)P_1.P_2\\
& < t^2+(1-t)^2 + 2t(1-t) \;(\mathrm{because}\;2t(1-t) > 0 \;\mathrm{and}\;P_1.P_2 < 1)\\
& = 1
\end{align}$$
A: We'll give a solution that works as well in hyperbolic geometry and perhaps in some other geometries.
Let $O$ the center of the circle, $A$, $B$ distinct points on the circle and $M$ a point on the segment $AB$. Consider the angles $\angle OMA  $ and $\angle OMB  $ summing up to $\pi$. Therefore, one of them has size $\ge \pi/2$. Assume it's $\angle OMA \ge \pi/2  $.  In the triangle $\Delta OMA$ the sum of the angles is $\ge \pi$. Therefore $\angle OAM < \pi/2 \le \angle OMA$ and therefore for the opposite sides we have $OM < OA$. 
