Constructing Tangents from an External Point Advanced. This lesson goes beyond core Geometry. You can skip it.

Problem. A circle centered at $O$ has radius $8$ cm. The given point $P$ is $20$ cm from $O$. Use a compass and unmarked straightedge to construct both tangent lines through $P$. Identify the center and radius of the second circle.

Solution. Use the given lengths to follow the drawing. Once the circle and point are in place, the tools locate each new point by intersection.

Step 1. Draw segment $OP$.

Step 2. Open the compass to a width greater than half of $OP$. From $O$, draw arcs to either side of $OP$. Keep the same opening and draw arcs from $P$ so the two pairs cross at $U$ and $V$. Draw line $UV$ and mark its intersection $M$ with $OP$. Since $U$ and $V$ are each equally far from $O$ and $P$, line $UV$ is the perpendicular bisector of $OP$.

Step 3. Center the compass at $M$ and open it to $O$. Draw the auxiliary circle, the extra circle that helps locate the contacts. Its radius is $10$ cm because $M$ bisects the $20$ cm segment. It also passes through $P$. Label its intersections with the original circle $T_1$ and $T_2$.

Step 4. Draw the lines through $P$ and $T_1$ and through $P$ and $T_2$. The short segments from $P$ to the contact points determine the two lines, which extend in both directions.

Problem. A point $T_1$ lies on a circle of radius $12$ mm centered at $O$. A point $P$ is $37$ mm from $O$ and $35$ mm from $T_1$. Decide whether line $PT_1$ is tangent to the circle at $T_1$. Justify your decision.

Solution. The longest side in triangle $OT_1P$ is $OP$. Compare its square with the sum of the other squares.

By the converse of the Pythagorean theorem, the angle opposite $OP$ is right. Therefore line $PT_1$ is perpendicular to the contact radius at $T_1$, so it is tangent there.

Problem. A completed construction for an outside point $P$ shows only the tangent at $T_1$, above the line $OP$ in the drawing. Describe a reflection that recovers the other contact point and explain why the reflected line is a different tangent through $P$.

Solution. Reflect across line $OP$. The center $O$ stays fixed, so the circle maps onto itself. The point $P$ also stays fixed, so the reflected line still passes through it.

Reflection preserves intersections. The reflected line meets the circle only at the image $T_2$ of $T_1$, so it is tangent. Since $T_1$ is above the mirror line, $T_2$ is below it and the contact points differ. A single tangent has just one contact point, so these are different tangents.

Problem. A circle has center $O$ and radius $11$ m. For each position of $P$, give the number of distinct tangent lines through $P$. In any case with just one line, describe how to construct it.

Case A. The distance $OP$ is $13$ m.

Case B. The distance $OP$ is $11$ m.

Case C. The distance $OP$ is $6$ m.

Solution. Case A. The distance exceeds the radius. The auxiliary circle gives two contact points, so there are two lines.

Case B. The distance equals the radius. Construct the perpendicular to $OP$ at $P$. This is the one tangent.

Case C. The distance is smaller than the radius. A line through $P$ crosses the circle in both directions from $P$, so there are no tangents.

Problem. A circle of radius $5$ cm has center $O$. The external point $P$ is $12$ cm from $O$. A second circle is drawn with center $P$ and compass opening $PO$. Can an intersection $X$ of these circles give a tangent line $PX$? Explain and give the center and compass opening that repair the construction.

Solution. The circle centered at $P$ makes $PX$ equal to $PO$. If $PX$ were tangent at $X$, triangle $OXP$ would be right at $X$. Then $PO$ would be its hypotenuse and would be longer than the leg $PX$. So no intersection found by that setup gives a tangent.

Put the compass at the midpoint $M$ of $OP$ and open it to $O$. The opening is $6$ cm. Now $O$ and $P$ are opposite ends of a diameter, so an intersection gives the needed right angle.