{{ item.displayTitle }}

No history yet!

equalizer

rate_review

{{ r.avatar.letter }}

{{ u.avatar.letter }}

+

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} Tangent lines have been defined and studied earlier in this course. In this lesson, methods for constructing a tangent line to a circle from an external point will be discussed.
### Catch-Up and Review

**Here is some recommended reading before getting started with this lesson.**

Consider a circle with center $O$ and point $P$ outside of the circle. Using a straightedge and compass, can you construct a tangent to the circle through the given point?

Think about the properties of tangents. How can you justify that the line you draw is indeed the tangent?

Given a circle and a point outside the circle, a compass and straightedge can be used to draw a tangent line from the point to the circle.

There are four steps to construct a tangent line.Draw $PO$

Mark Midpoint $M$ of $PO$

The point where $PO$ and its perpendicular bisector intersect is the midpoint of $PO.$

Draw a Circle with Center $M$ and Radius $MO$

Draw Tangents

The Inscribed Right Triangle Theorem can be used to justify why this construction works.

Consider $OA,$ a radius of $⊙O.$

In circle $M,$ $∠OAP$ is an inscribed angle on a diameter of $⊙M.$ Since an inscribed angle opposite the diameter is a right angle, $∠OAP$ is a right angle.

As can bee seen, $OA⊥PA.$ The radius of $⊙O$ is perpendicular to a line that passes through a point on the circle. Therefore, by the Tangent to Circle Theorem, $PA$ is tangent to the circle.

Kriz is learning a graphic program. By default, the program shows segment $AB$ and circle $O.$ The segment's endpoint $A$ can be moved anywhere outside of $⊙O.$ While endpoint $B$ can be moved anywhere. An eye-like shape appears on the screen when $AB$ is tangent to the circle. Give it a try!

Kriz can't quite place point $B$ in position to see the eye-like shape appear. Help Kriz out!

See solution.

How can a tangent line from a point outside of the given circle be constructed?

Since point $A$ is a point outside $⊙O,$ $B$ should be the point of tangency in order for $AB$ to be tangent to the circle. On the example shape, by extending $AB,$ it can be observed that $B$ is the point of tangency.

Constructing a tangent from an outer point will help locate the point of tangency for a tangent drawn from $A.$ Recall the steps in constructing a tangent.

- Connect the outer point and the center of the circle.
- Find midpoint $M$ of the segment drawn.
- Draw a circle with center $M$ and radius half the segment drawn.
- Draw tangent.

In this case, point $A$ is the outer point through which the tangent line is drawn. To get the example shape, move point A to the left as shown and then follow the steps.

As can be seen, the points where the circles intersect are the points of tangency. Therefore, point $B$ should be on these points.

The lines of symmetry of a circle are the lines that passes through the center of the circle.

Suppose that a tangent line drawn from an outer point intersects a circle at $A.$ When $A$ is reflected across the line that passes through $P$ and $O,$ its image will also be a point of tangency for another tangent.

Two tangent segments drawn from a common external point to the same circle are congruent.

For the above diagram, the following conditional statement holds true.

If $AB$ and $AC$ are tangent segments to $⊙O,$ then $AB≅AC.$

Consider two triangles.

- The triangle formed by the radius $OB,$ the segment $OA,$ and the tangent segment $AB.$
- The triangle formed by the radius $OC,$ the segment $OA,$ and the tangent segment $AC.$

These two triangles can be visualized in the diagram.

Note that $B$ and $C$ are points of tangency. Therefore, by the Tangent to Circle Theorem, $∠B$ and $∠C$ are right angles. Consequently, $△ABO$ and $△ACO$ are right triangles.

Because all radii of the same circle are congruent, it can be said that $OB$ and $OC$ are congruent. Moreover, $△ABO$ and $△ACO$ share the same hypotenuse $OA.$ By the Reflexive Property of Congruence, $OA$ is congruent to itself.

Combining all of this information, it can be said that the hypotenuse and one leg of $△ABO$ are congruent to the hypotenuse and the corresponding leg of $△ACO.$

$OB≅OCOA≅OA $

Therefore, by the Hypotenuse-Leg Theorem, $△ABO$ and $△ACO$ are congruent triangles. Since corresponding parts of congruent figures are congruent, it can be said that $AB$ and $AC$ are congruent.

$AB≅AC$

In the diagram, all three segments are tangent to circle $P.$

The points $D,$ $E,$ and $F$ are the points where the segments touch the circle. If $AB=12,$ $BC=10,$ and $CA=6,$ find $AE.$

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">A<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.05764em;\">E<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["4"]}}

Use the External Tangent Congruence Theorem.

From the graph, it can be seen that $AD$ and $AE$ are tangent segments with a common endpoint outside $⊙P.$ By the External Tangent Congruence Theorem, $AE$ and $AD$ are congruent.

Smilarly, $BE$ and $BF,$ and $CF$ and $CD$ are congruent tangent segments. $AD≅AEBE≅BFCF≅CD ⇔ AD=AEBE=BFCF=CD $ Using the Segment Addition Postulate and the given lengths, a system of equations with three equations and three unknowns can be written. $⎩⎪⎪⎨⎪⎪⎧ AE+BE=ABBF+CF=BCCD+AD=CA ⇒⎩⎪⎪⎨⎪⎪⎧ AE+BE=12BE+CF=10CF+AE=6 (I)(II)(III) $ To find $AE,$ the Elimination Method can be used. Start by multiplying the second equation by $-1.$ $-1(BE+CF)=-1(10)⇕-BE−CF=-10 $ Adding this equation to the first equation will eliminate $BE.$$AE+BE+(-BE−CF)=12+(-10)$

$AE−CF=2$

$CF+AE+(AE−CF)=6+(2)$

Simplify

RemovePar

Remove parentheses

$CF+AE+AE−CF=6+2$

AddSubFrac

Add and subtract fractions

$2AE=8$

DivEqn

$LHS/2=RHS/2$

$AE=4$

Imagine a superhero joining the Olympics to throw a hammer. An athlete would typically spin counterclockwise three or four (rarely five) times, then release the hammer. As viewed from above, the hammer travels on a path that is tangent to the circle created when the athlete spins. The diagram below shows the path of the superhero's hammer throw. See how the super hero fairs! Note, it is a not-to-scale drawing.

On August $30,$ $1986,$ in a packed stadium full of fans, Yuriy Sedykh set the world record with a throw of $86.74$ meters. Perhaps a throw farther than the superhero!

{{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

{{ exercise.headTitle }}

{{ 'ml-heading-exercise' | message }} {{ focusmode.exercise.exerciseName }}