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Task. Using the compass and straightedge, construct to a given circle the tangent lines through a given point outside the circle.
Solution. Let $O$ be the centre of the given circle and $P$ the given point. With $OP$ as diameter, draw the circle (see midpoint). If $A$ and $B$ are the points where this circle intersects the given circle, then by Thales' theorem, the angles $OAP$ and $OBP$ are right angles. According to the definition of the tangent of circle, the lines $AP$ and $BP$ are required tangents.
\begin{pspicture}(-5.5,-3)(5.5,3) \rput(-2.85,-0.1){$O$} \rput[linecolor=blue](+2.85,-0.1){$P$} \psdot(-2.6,0) \psdot(+2.6,0) \psline(-2.6,0)(2.6,0) \pscircle[linecolor=blue](-2.6,0){2} \psdot(0,0) \pscircle(0,0){2.6} \psdots(-1.8308,1.8462)(-1.8308,-1.8462) \psline[linestyle=dashed](-2.6,0)(-1.8308,+1.8462) \psline[linestyle=dashed](-2.6,0)(-1.8308,-1.8462) \psline[linestyle=dotted](-1.8308,+1.8462)(-1.8308,-1.8462) \rput(-1.8308,+2.2){$A$} \rput(-1.8308,-2.2){$B$} \psline[linecolor=blue](2.6,0)(-3,+2.3335) \psline[linecolor=blue](2.6,0)(-3,-2.3335) \rput(-5.5,-3){.} \rput(5.5,3){.} \end{pspicture}
The line segment $AB$ is
The convex angle $APB$ is called a tangent angle (or tangent-tangent angle) of the given circle and the convex angle $AOB$ the corresponding central angle. It is apparent that a tangent angle and the corresponding central angle are supplementary. The chord $AB$ is the tangent chord corresponding the tangent angle and the point $P$ (see equation of tangent
chord!).
The tangent angle is the angle of view of the line segment $AB$ from the point $P$ .
Note that if a circle is inscribed in a polygon, then the angles of the polygon are tangent angles of the circle and the centre of the circle is the common intersection point of the angle bisectors.
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