Tangent Lines & the Two-Tangent Theorem

A tangent line touches a circle at exactly one point, called the point of tangency. That single point of contact gives the tangent a special relationship with the circle: the radius drawn to the point of tangency is perpendicular to the tangent line.

That right angle is the reason tangent problems are really Pythagorean theorem problems. Draw the radius to the point of tangency and a right triangle appears. The second rule on this page, the Two-Tangent Theorem, says that two tangent segments drawn from the same outside point are congruent.

The radius meets the tangent at a right angle

If a line is tangent to a circle at point $T$ , then the radius $\overline{OT}$ is perpendicular to that line. So $m\angle OTP = 90^\circ$ for any point $P$ on the tangent line. This works in reverse as well: if a radius is perpendicular to a line at the point where the line meets the circle, that line is tangent.

The first move in almost every tangent problem is to draw that radius. Once it is there, the radius, the tangent segment, and the segment from the center to the outside point form a right triangle, with the segment from the center as the hypotenuse.

Finding a tangent length

Label the center $O$ , the point of tangency $T$ , and the outside point $P$ . The right angle sits at $T$ , so $\overline{OP}$ is the hypotenuse and the relationship is $OP^2 = r^2 + PT^2$ , where $r$ is the radius.

Which way you solve depends on what is missing. Looking for the distance to the center means adding the squares of the two legs. Looking for the tangent length or the radius means subtracting, because those are legs and $OP$ is the hypotenuse. Naming the hypotenuse before you compute keeps the addition and the subtraction straight.

Two tangents from the same point

Draw two tangent lines to a circle from a single outside point $P$ , touching the circle at $A$ and $B$ . The Two-Tangent Theorem says $PA = PB$ : the two tangent segments are congruent. Picture two zip lines running from the same tower to the same round roof.

That congruence turns into algebra fast. If the two tangent segments are given as expressions, set them equal and solve. There is an angle payoff too: the two radii and the two tangents close up a quadrilateral $OAPB$ , whose angles total $360^\circ$ . Two of those angles are the right angles at $A$ and $B$ , using $180^\circ$ , so the central angle $\angle AOB$ and the angle $\angle APB$ at the outside point are supplementary.

Worked examples

Example 1: find the tangent length

A circle has radius $3$ , and the outside point $P$ is $5$ units from the center. Find the tangent length $PT$ .

The radius meets the tangent at a right angle, so

OP

is the hypotenuse

OP^2 = r^2 + PT^2

Substitute the known lengths

5^2 = 3^2 + PT^2

Simplify

25 = 9 + PT^2

Subtract, then take the square root

PT^2 = 16, \; PT = 4

Answer: $PT = 4$

Example 2: find the distance to the center

A tangent segment from external point $P$ measures $12$ , and the circle's radius is $5$ . Find $OP$ .

The radius and the tangent segment are the two legs

OP^2 = r^2 + PT^2

Substitute

OP^2 = 5^2 + 12^2

Add

OP^2 = 25 + 144 = 169

Take the square root

OP = 13

Answer: $OP = 13$

Example 3: two tangents with algebraic lengths

$\overline{PA}$ and $\overline{PB}$ are tangent to circle $O$ from external point $P$ , with $PA = 2x + 5$ and $PB = 4x - 9$ . Find $x$ and the length of $\overline{PA}$ .

Two tangent segments from the same point are congruent

2x + 5 = 4x - 9

Subtract

2x

from both sides

5 = 2x - 9

Add

9

to both sides

14 = 2x

Divide by

2

x = 7

Substitute back to get the length

PA = 2(7) + 5 = 19

Answer: $x = 7$ and $PA = 19$ (and $PB = 4(7) - 9 = 19$ , which checks).

Example 4: the angle at the outside point

Two tangents from external point $P$ touch circle $O$ at $A$ and $B$ , and the central angle $\angle AOB$ measures $130^\circ$ . Find $m\angle APB$ .

The angles of quadrilateral

OAPB

total

360^\circ

m\angle O + m\angle A + m\angle P + m\angle B = 360

The angles at

A

and

B

are right angles

130 + 90 + m\angle P + 90 = 360

Combine the known angles

310 + m\angle P = 360

Subtract

m\angle P = 50

Answer: $m\angle APB = 50^\circ$ , the supplement of the $130^\circ$ central angle.

Try one yourself

Common questions

How do I check whether a line really is tangent?

Test the right angle with the Pythagorean theorem. If the radius is $8$ , the segment from the outside point to the touch point is $15$ , and the distance to the center is $17$ , then $8^2 + 15^2 = 289 = 17^2$ , so the angle is $90^\circ$ and the line is tangent. If the two sides do not match, the line is not tangent.

Does the Two-Tangent Theorem say the tangent lines are congruent?

It is about the segments, not the whole lines. The congruent pieces run from the outside point to each point of tangency, so $PA = PB$ . The lines themselves continue forever in both directions.

Why is the segment from the center always the hypotenuse?

Because the right angle is at the point of tangency, and the hypotenuse is always the side across from the right angle. That side is $\overline{OP}$ , running from the center to the outside point. So $OP$ is the longest of the three, and finding the radius or the tangent length means subtracting rather than adding.

Want the video version?

Allday Everyday Math has video lessons, practice, and an AI tutor for every topic, Pre-Algebra through Algebra 2.

Try it for $1