Updated 9.1.10-9.1.15

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@@ -87,11 +87,11 @@
</p>
<h3> Proof: </h3>
<p>Let the circles initially touch at point $A$, lying on the $Ox$ axis, where point $O$ is the center of the large circle. The coordinates of point $A$ are $ (kr, 0) $, where $ k = \frac{R}{r} $. Let us consider how the coordinates of point $A$, tied to the rolling circle ($A$ goes to $ A' $), change. Let the small circle roll so that its center moves from point $ C' $ to point $ C'' $ and rotates relative to point $ O $ by angle $ t $.</p>
−<p>First, it can be shown that the rotation of the small circle relative to its center (i.e. the angle between $ CA $ and $ C' A' $) is equal to $ t - kt = -(k-1)t $. Second, the coordinates of point $ C' $ will be as follows: $ (cos(t)(k-1)r, sin(t)(k-1)r) $. Then, knowing where the center of the rolling circle will go and at what angle it has turned relative to this center, we can write down the coordinates of point $ A' $:</p>
+<p>First, it can be shown that the rotation of the small circle relative to its center (i.e. the angle between $ CA $ and $ C' A' $) is equal to $ t - kt = -(k-1)t $. Second, the coordinates of point $ C' $ will be as follows: $ (cos(t)(k-1)r, \sin(t)(k-1)r) $. Then, knowing where the center of the rolling circle will go and at what angle it has turned relative to this center, we can write down the coordinates of point $ A' $:</p>
<p class="exp">
$ \begin{cases}
−x = cos(t)(k-1)r + cos((k-1)t)r \\
−y = sin(t)(k-1)r - sin((k-1)t)r
+x = \cos(t)(k-1)r + \cos((k-1)t)r \\
+y = \sin(t)(k-1)r - \sin((k-1)t)r
\end{cases} $
</p>
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