Tangents and Secants

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Learning goals

Classify a line using its intersections with a circle
Prove and apply radius-tangent perpendicularity and its converse
Find and justify equal tangent segment lengths from one external point
Distinguish internal and external contact between two circles
Count common tangents using circle positions and the centers' sides

Count intersections first

A secant is a line meeting a circle at two distinct points. A tangent is a line meeting it at exactly one point, called the point of tangency or contact point. A line with no intersection is neither. These definitions concern the entire line, even when a drawing shows only part of it.

For example, take a circle of radius $5$ units. A line $6$ units from its center misses it. A line $5$ units from the center touches once. A line $3$ units from the center cuts twice: the perpendicular from the center leaves a right triangle with hypotenuse $5$ and one leg $3$ on each side of its foot.

The same circle appears in all three panels. Count intersections with the complete line.

To see why these are the only possibilities, let $H$ be the perpendicular foot from center $O$ to a line. Every other point $X$ on the line is farther from $O$ than $H$ is. If $OH$ exceeds the radius, no point can be on the circle. If $OH$ equals the radius, only $H$ is on it. If $OH$ is smaller, Pythagoras gives one equal positive distance from $H$ in each direction along the line, hence two circle points. This includes a line through $O$ , whose intersections are the endpoints of a diameter.

Worked example 1 Classify three lines without trusting the sketch

Problem. A circle has radius $7$ cm. The perpendicular distances from its center to lines $a$ , $b$ , and $c$ are $7$ cm, $9$ cm, and $2$ cm, respectively. Classify each line and give its number of circle intersections.

Solution. For line $a$ , the distance equals the radius. It is a tangent with one intersection.

For line $b$ , the distance is greater than the radius. It misses the circle and has no intersections.

For line $c$ , the distance is smaller than the radius. It is a secant with two intersections.

The three configurations illustrate the classifications in the solution.

Check your understanding

A circle has radius $6$ units. A line is $2$ units from its center. How does the line meet the circle?

A It misses the circle B It is tangent C It is a secant D It lies entirely on the circle

A tangent and its contact radius

A tangent meets its contact radius at a right angle. The contact point matters: the tangent need not be perpendicular to another radius.

A tangent is perpendicular to its contact radius#

First consider a radius- $13$ circle and a line whose perpendicular distance from the center is $5$ . Pythagoras gives $12$ units from the foot $H$ to either intersection. Reflecting one intersection across the perpendicular through $H$ gives the other. A line whose foot differs from its circle intersection therefore cannot touch only once.

This is an actual secant. Reflection explains why a line with its perpendicular foot away from a circle point has a second intersection.

Now let a tangent meet a positive-radius circle at $T$ , and let $H$ be the perpendicular foot from $O$ to that line. Suppose $H$ differs from $T$ . Reflect in the line through $O$ perpendicular to the tangent. This reflection fixes $O$ , carries the tangent line onto itself, and sends $T$ to a different point $U$ on that line. Distance preservation gives a second point of the circle because $OU$ equals radius $OT$ . This contradicts the definition of a tangent. Thus $H$ is $T$ , and the radius is perpendicular to the tangent. If $H$ is $O$ , the same reflection still works; it exchanges the two ends of a diameter. Nothing depended on radius $13$ .

The reverse claim is useful when tangency is not given.

A perpendicular at a radius endpoint is a tangent#

Take a radius of $6$ units and a line perpendicular to it at its endpoint $T$ on the circle. A point $X$ on the line, $8$ units from $T$ , is $10$ units from the center by Pythagoras, so it is outside. The same comparison works for every point other than $T$ .

Given the right angle at T, every other line point X is farther from O than radius OT.

In general, let $T$ be on a circle with center $O$ and positive radius $r$ . A line through $T$ is perpendicular to $OT$ . For any other point $X$ on that line, the right triangle gives

\begin{aligned} OX^2&=r^2+TX^2,\\ OX^2&>r^2. \end{aligned}

Hence $X$ is outside the circle, while $T$ is on it. The line meets the circle exactly once and is a tangent. The numerical values above did not matter; the positive extra square did.

Together these results say: a line through a point on a circle is tangent exactly when it is perpendicular to the radius at that point. Perpendicularity somewhere else on the radius is not enough.

Worked example 2 Find a finite tangent segment

Problem. A circle centered at $O$ has radius $9$ cm. Point $P$ is $15$ cm from $O$ , and line $PT$ is tangent at $T$ . Find the length of segment $PT$ .

The stated tangent gives the right angle. The unknown segment has no given numerical label.

Solution. Radius $OT$ is perpendicular to tangent $PT$ , so $OP$ is the hypotenuse of right triangle $OTP$ . Use the Pythagorean theorem:

\begin{aligned} PT^2+9^2&=15^2,\\ PT^2&=144,\\ PT&=12\text{ cm}. \end{aligned}

Choose the positive root because $PT$ is a length. We have measured a finite segment from $P$ to $T$ , not an infinite line.

Check your understanding

Point $T$ is on a circle centered at $O$ . A line through $T$ is perpendicular to $OT$ . Why is it tangent?

A Every other point of the line is farther from O than T is B It passes through the center C It is perpendicular to every chord D It must cross the circle a second time

Two tangent segments from one external point

A strict external point is farther from the center than the radius. From such a point $P$ , a circle has two distinct tangent lines, touching at $T$ and $U$ . The two finite segments $PT$ and $PU$ have equal lengths.

For example, if the radius is $8$ and $OP$ is $10$ , both right triangles have the same hypotenuse $10$ and a radius leg $8$ . Pythagoras gives a tangent length of $6$ in each triangle.

Tangent segments from a common external point are equal#

The numerical instance has two right triangles with common hypotenuse OP. Each finite tangent segment is 16 units.

For the numerical case, both triangles are right at their contact points; each has hypotenuse $20$ and a leg $12$ . The HL congruence criterion makes the triangles congruent, so their remaining sides both have length $16$ .

For any positive-radius circle and strict external point $P$ , join $OT$ , $OU$ , and $OP$ . The radius-tangent theorem gives right angles at $T$ and $U$ . The radius legs are equal, and $OP$ is the common hypotenuse. HL gives

\triangle OTP\cong\triangle OUP.

Corresponding sides therefore give

PT=PU.

The proof did not depend on $12$ or $20$ . It does depend on both segments being tangent to the same circle from the same strict external point.

Compare the two tangent segments

Point $P$ must stay strictly outside for two distinct positive tangent segments. A point on the circle has one tangent line at that point; it does not give two positive contact segments. A point strictly inside has no tangent line through it, since every tangent line’s closest distance from $O$ is the radius. The familiar two-tangent picture therefore has a real boundary condition.

Move $P$ through the six external positions. Predict whether $PA$ and $PB$ will stay equal, then compare the displayed lengths.

Two tangent segments from one external point

Point P

The radius is 4 units. OP = 6 units. PA ≈ 4.47 units. PB ≈ 4.47 units.

P stays outside a circle of radius 4. The perpendicular radius contacts distinguish the tangents from secants, and matching ticks pair PA and PB.

Worked example 3 Use equality before solving

Problem. The tangent segments $PT$ and $PU$ from the same strict external point touch the same circle at $T$ and $U$ . Their lengths, in centimeters, are $3x+2$ and $5x-10$ , respectively. Find $x$ and both segment lengths.

The two algebraic labels are given segment lengths. The picture is schematic.

Solution. Both tangent segments meet the theorem’s conditions, so set their lengths equal:

\begin{aligned} 3x+2&=5x-10,\\ 12&=2x,\\ x&=6. \end{aligned}

Substitute into each expression:

\begin{aligned} PT&=20\text{ cm},\\ PU&=20\text{ cm}. \end{aligned}

Both are positive, as the geometry requires.

Check your understanding

A circle has radius $8$ cm, and $OP$ is $17$ cm. Tangents from $P$ touch at $T$ and $U$ . What is the length of each tangent segment?

9

cm B

15

cm C

19

cm D

25

When two circles touch

Two distinct circles are tangent circles when they share exactly one point. At that point their tangent lines agree, and their contact radii lie on the same line. If the radii were not collinear, reflection in the line of centers would give a second shared point.

The two contact configurations are separate panels. The tangent at T is perpendicular to the common center line.

With radii $5$ and $2$ units, external contact puts the contact point between the centers and gives center distance $7$ . Internal contact puts the smaller center between the larger center and the contact point, giving distance $3$ .

Let the larger radius be $R$ , the smaller be $r$ , and the center distance be $d$ . Both radii are positive.

External contact has the circles outside one another:

d=R+r.

Internal contact has one circle inside the other and requires unequal radii:

d=R-r.

Equal radii with the same center produce coincident circles, which share the whole circle, not a single contact point.

Worked example 4 Distinguish two kinds of contact

Problem. Two circles have radii $10$ cm and $4$ cm. Classify their contact in each case.

Case A: their centers are $14$ cm apart.

Case B: their centers are $6$ cm apart.

Solution. Case A: the sum of the radii is $14$ cm, so the circles are externally tangent. The contact point lies between the centers.

Case B: the difference of the radii is $6$ cm, so the circles are internally tangent. The smaller circle lies inside the larger, touching it once.

These schematic contact positions illustrate the two cases. Use the given radii and center distances.

Check your understanding

Two circles have radii $9$ cm and $5$ cm, and their centers are $4$ cm apart. Which description fits?

A Externally tangent B Intersecting twice C Separate with neither inside the other D Internally tangent

Common tangents depend on both centers

A common tangent is one line tangent to both circles. It is a common external tangent when the two centers lie on the same side of that line. It is a common internal tangent when the centers lie on opposite sides. These names concern sides of a line; they do not say whether the circles touch externally or internally.

Classify the common tangent by the centers, not by whether the line appears between the circles.

For radii $5$ and $3$ , a common external tangent leaves a perpendicular radius difference of $2$ between the centers. An internal tangent leaves a perpendicular separation of $8$ , the sum of the radii. The center distance must be at least that perpendicular separation. For example, center distance $6$ permits external tangents but is too short for an internal one.

Here is why the thresholds also give the counts. Assume first that the centers are distinct. For a common external tangent, both contact radii point toward the same side. The perpendicular part of the center separation is the radius difference. If the center distance is larger than this difference, a right triangle has a positive along-line leg. Its two mirror placements give two external tangents. At equality that leg disappears, leaving one. Below the difference no such triangle is possible.

For a common internal tangent, the contact radii point in opposite directions. Replace the difference with the radius sum in the same argument. Larger center distance gives two internal tangents; equality gives one; smaller distance gives none. The two mirror choices exhaust the possibilities because a right triangle with the fixed hypotenuse and perpendicular leg has only those two placements. For equal radii and distinct centers, the external tangents are the two parallels to the line of centers. This is the zero-difference case.

Each drawn line is one distinct common tangent. The circles use a common length scale across the panels.

Intersecting circles retain two external tangents; internal contact leaves one.

Strict containment leaves no line that is tangent to both circles.

If the centers coincide but the radii differ, a line cannot be at two different perpendicular distances from the same center, so there is no common tangent. If both center and radius agree, the circles coincide and every tangent to one is tangent to the other.

The solid and dashed circles coincide. The single representative line stands for infinitely many possible common tangents.

The complete count uses $R\ge r>0$ and counts distinct lines. In the table, circles described as nested have no contact.

Configuration	Center distance	External	Internal	Total
Separate, neither inside the other	$d>R+r$	2	2	4
Externally tangent	$d=R+r$	2	1	3
Intersecting twice	$R-r<d<R+r$	2	0	2
Internally tangent, unequal radii	$d=R-r>0$	1	0	1
Nested, distinct centers	$0<d<R-r$	0	0	0
Concentric, unequal radii	$d=0$	0	0	0
Coincident, equal radii	$d=0$	Infinitely many	0	Infinitely many

At external contact the two internal tangents have merged into the one line through the shared contact. At internal contact the two external tangents have merged. Count that line once, even though it touches both circles at the same point.

Worked example 5 Count the tangent lines at a boundary

Problem. Two circles have radii $6$ cm and $2$ cm. Give the number of common external tangents, common internal tangents, and all common tangents in each case.

Case A: center distance $9$ cm.

Case B: center distance $8$ cm.

Solution. Case A: $9$ exceeds the radius sum, $8$ . The circles are separate. There are two external and two internal common tangents, for four distinct lines.

Case B: the center distance equals the radius sum. The circles touch externally. The two external tangents remain, but there is just one internal tangent at the contact point, for three distinct lines.

These schematic tangent configurations illustrate the two counts in the solution.

Check your understanding

Two circles have the same center and the same positive radius. How many common tangent lines do they have?

A 0 B 1 C 2 D Infinitely many

Tangents and Secants

Learning goals

Count intersections first

A tangent and its contact radius

Two tangent segments from one external point

Compare the two tangent segments

When two circles touch

Common tangents depend on both centers

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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