How do I know whether to add or subtract the arcs?

Look only at the vertex. Vertex inside the circle means add the two intercepted arcs and halve; vertex outside means subtract the near arc from the far arc and halve; vertex on the circle means halve the single intercepted arc. If subtraction gives a negative number, you labeled far and near backwards.

Why is a tangent perpendicular to the radius at the point of tangency?

The shortest distance from the center to a line is along the perpendicular. If the radius were not perpendicular, the perpendicular distance from the center to the line would be less than the radius, so the line would dip inside the circle and cross it twice — making it a secant, not a tangent.

Does the chord-tangent angle formula ever conflict with the inscribed angle formula?

No, they agree. A chord-tangent angle is the limiting case of an inscribed angle as one endpoint slides toward the vertex. Both equal half the intercepted arc, so an inscribed angle and a chord-tangent angle cutting off the same arc have the same measure.

Are two tangent segments from an outside point always equal, even if the circle is huge or the point is far away?

Yes. The proof uses HL congruence on the two right triangles formed by the radii, the tangent segments, and the shared segment from the point to the center, and nothing in that argument depends on the size of the circle or the distance. As long as the point lies outside the circle, both tangent segments have length sqrt(OP^2 - r^2).

GEOM-9.3

Tangents, Secants & Angle Measures

Master tangent-radius perpendicularity, congruent tangent segments, and the half-sum and half-difference rules for angles formed by chords, secants, and tangents.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Tangents, Secants & Angle Measures, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Tangents, Secants & Angle Measures free

What this lesson covers

Every angle you can build with a circle comes down to one question: where is the vertex? If the vertex sits at the center, the angle equals its arc. If it sits on the circle, the angle is half its arc — you saw that with inscribed angles. This lesson finishes the picture by adding tangent lines and by moving the vertex inside and outside the circle.

You will learn two structural facts about tangents (a tangent is perpendicular to the radius at the point of tangency, and two tangent segments from the same outside point are congruent), plus three angle formulas that cover every remaining case. Once you can classify a diagram by vertex location, the rest is arithmetic. The hardest part for most students is not the formulas but deciding which arcs a given angle actually intercepts, so we spend real time on that.

Tangents, Radii, and Congruent Tangent Segments

A tangent is a line that touches a circle at exactly one point, called the point of tangency. A secant is a line that cuts through a circle at two points. A chord is the segment between those two intersection points.

The first key theorem: a tangent line is perpendicular to the radius drawn to the point of tangency. If line

PA

is tangent to circle

O

A

, then

\overline{OA} \perp \overline{PA}

, so

m\angle OAP = 90^\circ

. The converse is also true, which is how you prove a line is tangent: show the radius meets it at a right angle.

That right angle is useful because it drops a right triangle into the picture. In right triangle

OAP

, the legs are the radius

OA

and the tangent segment

AP

, and the hypotenuse is

OP

, the distance from the outside point to the center. So

OA^2 + AP^2 = OP^2

.

The second key theorem: if two tangent segments are drawn to a circle from the same external point, they are congruent. If

\overline{PA}

and

\overline{PB}

are both tangent to circle

O

, then

PA = PB

. The quick proof uses the two right triangles

OAP

and

OBP

: they share hypotenuse

OP

, have congruent legs

OA \cong OB

(radii), and each contains a right angle, so they are congruent by HL.

A consequence students often need:

\overrightarrow{PO}

bisects

\angle APB

, and quadrilateral

OAPB

has two right angles at

A

and

B

, so

m\angle AOB + m\angle APB = 180^\circ

Locating the Vertex: Three Angle Rules

Once tangents are in play, every angle formed by chords, secants, and tangents falls into one of three cases based on the vertex.

Vertex location	Formed by	Angle measure
On the circle	Two chords, or a chord and a tangent	Half the one intercepted arc
Inside the circle	Two chords crossing	Half the sum of the two intercepted arcs
Outside the circle	Two secants, two tangents, or one of each	Half the difference of the two intercepted arcs

Vertex on the circle: the chord-tangent (or tangent-chord) angle equals half the arc it cuts off. If

\overline{AB}

is a chord and

\overrightarrow{AT}

is tangent at

A

, then

m\angle TAB = \tfrac{1}{2}m\widehat{AB}

, using the arc that lies inside the angle. Notice this matches the inscribed angle rule — a tangent is the limiting position of a secant.

Vertex inside: two chords meet at an interior point, forming vertical angle pairs. Each angle is half the sum of its own intercepted arc and the arc intercepted by its vertical angle:

m\angle 1 = \tfrac{1}{2}\left(m\widehat{AB} + m\widehat{CD}\right)

Vertex outside: the two sides cut the circle so that one arc is far from the vertex and one is near. Then

m\angle P = \tfrac{1}{2}\left(m\widehat{\text{far}} - m\widehat{\text{near}}\right)

This single formula covers secant-secant, secant-tangent, and tangent-tangent; only the way you name the arcs changes. For two tangents, the near and far arcs together make the whole circle, so

m\angle P = \tfrac{1}{2}(360 - 2x) = 180 - x

where

x

is the near arc.

Choosing the Right Arcs

The formulas are short; identifying arcs is where the work is. Three habits prevent most errors.

First, for an interior angle, do not use the arcs that the angle appears to "point between" on one side only. Extend both chords mentally: the angle and its vertical partner each open toward one arc, and you add those two arcs. A frequent wrong move is adding an arc to itself or grabbing an adjacent arc that neither ray intercepts.

Second, for an exterior angle, the far arc is the one whose endpoints are the two points farther from the vertex; the near arc's endpoints are the two points closer. Subtract near from far, never the reverse — a negative answer is a signal you flipped them. Also, the far arc alone is not the answer, and the difference before halving is not the answer.

Third, remember the whole circle is

360^\circ

. Many problems give you two arcs and expect you to find a third by subtraction before applying a formula.

A useful sanity check ties the three cases together. Keep the same two arcs and slide the vertex outward: at an interior point the angle is half a sum (large), on the circle it is half one arc, and outside it is half a difference (small). So an exterior angle is always smaller than either inscribed angle on the same arcs. If your exterior angle comes out larger than half the far arc, recheck.

Finally, tangent problems often mix in the right angle from the radius. When a diagram shows a radius drawn to a point of tangency, mark the

90^\circ

immediately — many multi-step problems are really a triangle angle-sum problem in disguise.

Putting Tangents and Angles Together in Proofs

Most homework problems in this section are two-step: use a tangent property to unlock a length or a right angle, then use an arc formula to finish, or the reverse.

A typical chain looks like this. Tangent segments

\overline{PA}

and

\overline{PB}

are drawn from external point

P

. You are told

m\angle APB = 46^\circ

. Because

OAPB

has right angles at

A

and

B

, the central angle

\angle AOB

measures

180 - 46 = 134^\circ

, so the near arc

\widehat{AB} = 134^\circ

and the far arc is

360 - 134 = 226^\circ

. Checking with the exterior formula:

\tfrac{1}{2}(226 - 134) = \tfrac{1}{2}(92) = 46^\circ

. The two routes agree, which is a good way to verify your arc labeling.

Algebraic versions are common too. If tangent segments are given as

PA = 3x + 4

and

PB = 5x - 6

, congruence gives

3x + 4 = 5x - 6

, so

x = 5

and each segment is 19 units.

When writing a proof, name the theorem you use: "tangent perpendicular to radius at the point of tangency," "tangent segments from a common external point are congruent," "the measure of an angle formed by a chord and a tangent is half the intercepted arc." Teachers look for that justification, not just the number.

One caution: the congruent-tangent-segment theorem applies only to segments from an external point to the points of tangency. It says nothing about chords or about secant segments, which follow different (product) relationships you may meet later.

Key terms

Tangent line.: A line in the plane of a circle that intersects the circle at exactly one point, the point of tangency.
Secant line.: A line that intersects a circle at exactly two points; the segment between those points is a chord.
Point of tangency.: The single point where a tangent line touches the circle; the radius drawn to it is perpendicular to the tangent.
Tangent segment.: The segment from an external point to a point of tangency. Two such segments from the same external point are congruent.
Chord-tangent angle.: An angle whose vertex is on the circle, formed by a chord and a tangent; its measure is half the intercepted arc.
Intercepted arc.: The arc lying in the interior of an angle, with endpoints on the angle's sides.
Far arc and near arc.: For a vertex outside the circle, the far arc has endpoints farther from the vertex and the near arc has endpoints closer; the angle is half their difference.
Half-sum rule.: For two chords intersecting inside a circle, each angle equals half the sum of its intercepted arc and the arc intercepted by its vertical angle.

Worked example

From external point

P

, line

PA

is tangent to circle

O

A

, and a secant from

P

passes through the circle hitting it first at

B

and then at

C

. The arcs are

m\widehat{AB} = 70^\circ

(the arc not containing

C

) and

m\widehat{AC} = 190^\circ

(the arc not containing

B

). Find (a)

m\widehat{BC}

, (b)

m\angle P

, (c)

m\angle PAB

, and (d)

m\angle OAP

Start with the whole circle. Points

A

B

C

divide it into three arcs that must total

360^\circ

m\widehat{BC} = 360 - 70 - 190 = 100^\circ

(b) The vertex

P

is outside the circle, so use the half-difference rule. The sides of the angle are the tangent ray through

A

and the secant ray through

B

and

C

. The points farther from

P

are

A

and

C

, so the far arc is

\widehat{AC} = 190^\circ

; the points nearer

P

are

A

and

B

, so the near arc is

\widehat{AB} = 70^\circ

. Then

m\angle P = \tfrac{1}{2}(190 - 70) = \tfrac{1}{2}(120) = 60^\circ

Note that

A

appears in both arcs — that is normal for a tangent-secant angle, since the tangent touches at only one point.

(c)

\angle PAB

has its vertex on the circle and is formed by tangent

\overrightarrow{AP}

and chord

\overline{AB}

. The arc inside that angle is

\widehat{AB}

, so

m\angle PAB = \tfrac{1}{2}(70) = 35^\circ

(d) The radius

\overline{OA}

meets the tangent at the point of tangency, so

m\angle OAP = 90^\circ

.

Check part (b) a second way: in triangle

PAB

, the exterior angle at

B

equals the inscribed angle

\angle ABP

's supplement. The inscribed angle

\angle ACB

intercepts

\widehat{AB} = 70^\circ

, giving

35^\circ

, and the inscribed angle

\angle BAC

intercepts

\widehat{BC} = 100^\circ

, giving

50^\circ

. Angle

ABC

intercepts

\widehat{AC} = 190^\circ

, so it measures

95^\circ

and

m\angle ABP = 85^\circ

. In triangle

PAB

180 - 85 - 35 = 60^\circ

. It matches.

Practice questions

Two chords

\overline{AC}

and

\overline{BD}

intersect inside a circle at point

E

. If

m\widehat{AB} = 84^\circ

and

m\widehat{CD} = 36^\circ

, what is

m\angle AEB

$24^\circ$
$48^\circ$
$60^\circ$
$120^\circ$

Answer: $60^\circ$

The vertex is inside the circle, so the angle is half the sum of the arc it intercepts and the arc intercepted by its vertical angle:

m\angle AEB = \tfrac{1}{2}(84 + 36) = \tfrac{1}{2}(120) = 60^\circ

. The choice

24^\circ

comes from subtracting instead of adding (that is the outside-vertex rule, which does not apply here), and

120^\circ

comes from forgetting to take half.

Segments

\overline{PA}

and

\overline{PB}

are tangent to circle

O

A

and

B

. Given

PA = 4x - 3

PB = 2x + 9

, and

m\widehat{AB} = 118^\circ

(the minor arc), find

x

, the length

PA

, and

m\angle APB

. Justify each step.

Answer: $x = 6$ , $PA = 21$ units, and $m\angle APB = 62^\circ$ .

Because two tangent segments drawn from the same external point are congruent,

4x - 3 = 2x + 9

, so

2x = 12

and

x = 6

. Substituting,

PA = 4(6) - 3 = 21

, and the check

PB = 2(6) + 9 = 21

confirms it. For the angle, the vertex is outside the circle, so use the half-difference rule. The near arc is the minor arc

\widehat{AB} = 118^\circ

and the far arc is

360 - 118 = 242^\circ

. Then

m\angle APB = \tfrac{1}{2}(242 - 118) = \tfrac{1}{2}(124) = 62^\circ

. You can verify with the quadrilateral

OAPB

: the radii meet the tangents at right angles, so the four angles give

90 + 90 + 118 + m\angle APB = 360

, again

62^\circ

A tangent touches circle

O

at point

T

, and chord

\overline{TR}

is drawn so that the chord-tangent angle on one side measures

58^\circ

. Find the measure of the arc

\widehat{TR}

intercepted by that angle, and the measure of the chord-tangent angle on the other side of the tangent line.

Answer: The intercepted arc is $116^\circ$ , and the angle on the other side is $122^\circ$ .

A chord-tangent angle equals half its intercepted arc, so

58 = \tfrac{1}{2}m\widehat{TR}

, giving

m\widehat{TR} = 116^\circ

. The two chord-tangent angles at

T

form a linear pair, so the other one measures

180 - 58 = 122^\circ

. Consistency check: that larger angle intercepts the major arc

\widehat{TR} = 360 - 116 = 244^\circ

, and

\tfrac{1}{2}(244) = 122^\circ

. This is why you must use the arc lying inside the angle you are working with.

FAQ

How do I know whether to add or subtract the arcs?: Look only at the vertex. Vertex inside the circle means add the two intercepted arcs and halve; vertex outside means subtract the near arc from the far arc and halve; vertex on the circle means halve the single intercepted arc. If subtraction gives a negative number, you labeled far and near backwards.
Why is a tangent perpendicular to the radius at the point of tangency?: The shortest distance from the center to a line is along the perpendicular. If the radius were not perpendicular, the perpendicular distance from the center to the line would be less than the radius, so the line would dip inside the circle and cross it twice — making it a secant, not a tangent.
Does the chord-tangent angle formula ever conflict with the inscribed angle formula?: No, they agree. A chord-tangent angle is the limiting case of an inscribed angle as one endpoint slides toward the vertex. Both equal half the intercepted arc, so an inscribed angle and a chord-tangent angle cutting off the same arc have the same measure.
Are two tangent segments from an outside point always equal, even if the circle is huge or the point is far away?: Yes. The proof uses HL congruence on the two right triangles formed by the radii, the tangent segments, and the shared segment from the point to the center, and nothing in that argument depends on the size of the circle or the distance. As long as the point lies outside the circle, both tangent segments have length $\sqrt{OP^2 - r^2}$ .

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