Geometry · G.C.A.2 · G.C.A.4

Proving a tangent is perpendicular to the radius

A line touches a circle at exactly one point. Why does that force the radius to meet the line at a right angle?

You need

Indirect proofGeometry Tangent propertiesGeometry

Where you are 3 in Proofs about circles

Geometry11

Geometric proof18

Proofs about circles3

Proving the inscribed angle theorem

Proving a tangent is perpendicular to the radius

Proving the power of a point relationships

When you reach for it

A line is tangent to a circle at a point P when P is the only point the line and the circle share. Nothing in that sentence mentions a right angle. Yet in almost every problem about tangents the first thing written down is that the radius to the point of contact is perpendicular to the line.

Figure 1The line touches the circle only at P, and the radius OP meets it at a right angle.

This lesson proves that right angle from the one-point definition, using nothing but a perpendicular from the center, two congruent triangles and the Pythagorean theorem. It also proves the converse, so the statement can be used in both directions:

Theorem. Let \ell be a line, and let P be a point of \ell that lies on a circle with center O. Then \ell is tangent to the circle at P if and only if OP \perp \ell.

Reach for the first half, tangent gives perpendicular, when a figure shows a tangent and a proof or calculation needs a right angle or a right triangle. Reach for the second half, perpendicular gives tangent, when you have to show that a given line really is a tangent. The proof of the first half is also a clear example of indirect proof: assume the opposite, and follow it until it contradicts the definition.

The lesson Tangent properties uses this statement freely. Here is why it is true.

The procedure

An “if and only if” is two claims, and each needs its own proof. The claims are the theorem and its converse: tangent gives perpendicular, and perpendicular gives tangent.

Tangent gives perpendicular

Suppose \ell is tangent to the circle at P. The goal is OP \perp \ell.

Assume the opposite. Suppose OP is not perpendicular to \ell. The plan is to show that then \ell has a second point on the circle, which the word “tangent” forbids.

Rule out a line through the center. If O were on \ell, then \ell would contain P and also the point on the far side of O at distance r from it, two different points of the circle. So O is not on \ell.

Drop the perpendicular from the center. Because O is not on \ell, there is exactly one perpendicular from O to \ell. Call its foot Q, so OQ \perp \ell. The point Q is not P: OQ is perpendicular to \ell and, by the assumption, OP is not.

Figure 2If OP is slanted, the mirror image P' of P across the perpendicular OQ is a second point of the circle on the line.

Reflect P across the foot. Let P' be the point of \ell on the opposite side of Q from P with QP' = QP. Since Q \neq P, the length QP is positive, so P' is a different point from P.

Compare two right triangles. In \triangle OQP and \triangle OQP':

OQ = OQ, the same segment,
\angle OQP = \angle OQP' = 90^\circ, since OQ \perp \ell,
QP = QP', by the choice of P'.

Two sides and the angle between them agree, so the triangles are congruent, and corresponding sides give OP' = OP. But OP is a radius, so OP' = r as well.

Read off the contradiction. P' is on \ell, and it is at distance r from O, so it is on the circle. Now \ell meets the circle at P and at P', two different points. A tangent meets the circle at one point only. The assumption cannot stand, and

OP \perp \ell .

Perpendicular gives tangent

Suppose now that P is on the circle, \ell passes through P, and OP \perp \ell. The goal is to show that P is the only point of \ell on the circle.

Take any other point. Let X be any point of \ell other than P. Then OPX is a triangle with a right angle at P, and OX is its hypotenuse.

Figure 3When OP \perp \ell, the segment OX is the hypotenuse of a right triangle, so it is longer than the radius OP.

Use the Pythagorean theorem.

OX^2 = OP^2 + PX^2 = r^2 + PX^2 .

Compare with the radius. PX > 0, so OX^2 > r^2, and therefore OX > r. The point X is farther from the center than a point of the circle can be, so X is outside the circle and is not on it. Since X was any point other than P, the only common point is P, and \ell is tangent to the circle at P.

This half says more than it needed to. Every point of \ell other than P is outside the circle, so the radius OP is the shortest segment from O to \ell.

A second route: the shortest path

The shortest-path picture gives a second proof of the first half. If \ell is tangent at P, then every other point X of \ell is outside the circle. Here is why. Suppose some X \neq P were inside, with OX < r. Walk along \ell from X away from P. The distance from O starts below r and grows without bound, and it changes without jumping, so it passes through r at some point. That point is on the circle and on \ell, and it is on the other side of X from P, so it is not P. A second common point is again impossible.

So OX > r = OP for every X \neq P on \ell: P is the point of \ell nearest to O. The nearest point of a line to a given point is the foot of the perpendicular, as shown in Distance from a point to a line, so OP \perp \ell. The first proof needs only congruent triangles; this one needs that fact about shortest distance, and each is worth knowing.

What the theorem lets you write

In a proof, the reason for “OP \perp \ell” when \ell is tangent at P is: a tangent is perpendicular to the radius drawn to the point of contact. The reason for “\ell is tangent at P” is: a line through a point of a circle that is perpendicular to the radius at that point is tangent to the circle. And because the two halves are converses, a line through P that is not perpendicular to OP is not tangent, a line that cuts the circle a second time.

Worked examples

Example 1

An angle between a tangent and a chord

The line \ell is tangent at P to a circle with center O. The chord PA has \angle POA = 70^\circ. Find the angle between the chord PA and the half of \ell that starts at P on the same side of OP as A.

Solution

OP and OA are radii, so \triangle OPA is isosceles with apex O. Its base angles are equal, and the three angles add to 180^\circ:

\angle OPA = \frac{180^\circ - 70^\circ}{2} = 55^\circ .

The tangent is perpendicular to the radius OP, so the angle between \ell and OP is 90^\circ. The chord PA splits that right angle into \angle OPA and the angle we want:

90^\circ - 55^\circ = 35^\circ .

The angle between the tangent and the chord is 35^\circ, which is half of 70^\circ. That is no accident; it is the subject of Tangent chord angle.

Example 2

The length of a tangent

A circle has center O and radius 5. From a point A with OA = 13, the tangent AB touches the circle at B. Find AB.

Solution

Figure 4The tangent from A makes a right triangle with the radius and OA.

AB is tangent at B, so OB \perp AB and \triangle OBA has a right angle at B. Its hypotenuse is OA = 13 and one leg is the radius OB = 5. Let x = AB. By the Pythagorean theorem,

\begin{aligned} x^2 + 5^2 &= 13^2 \\ x^2 &= 169 - 25 = 144 \\ x &= 12 . \end{aligned}

The length is positive, so only x = 12 is possible. The right angle is what makes the Pythagorean theorem available at all, which is why the tangent theorem comes first.

Example 3

Is this line a tangent?

A circle has center O and radius 10. The point P is on the circle, and \ell is a line through P. A point X on \ell has PX = 24 and OX = 26. Is \ell tangent at P? What if OX = 25 instead?

Solution

The segment OP is a radius, so OP = 10. Test whether \triangle OPX has a right angle at P using the converse of the Pythagorean theorem:

10^2 + 24^2 = 100 + 576 = 676 = 26^2 .

The squares of the two shorter sides add to the square of the longest, so the angle at P is a right angle. Then OP \perp \ell, and by the second half of the theorem \ell is tangent at P.

With OX = 25: 10^2 + 24^2 = 676 but 25^2 = 625. They are not equal, so the angle at P is not a right angle and OP is not perpendicular to \ell. By the first half of the theorem, a tangent at P would have to be perpendicular to OP, so \ell is not tangent. It crosses the circle a second time.

Example 4

The equation of a tangent

Find the equation of the line tangent at P(3, 4) to the circle x^2 + y^2 = 25, and confirm that it meets the circle only once.

Solution

P is on the circle, since 3^2 + 4^2 = 25. The center is O(0, 0), so the radius OP has slope \displaystyle \frac{4}{3}. The tangent is perpendicular to it, and perpendicular lines have slopes whose product is -1, so the tangent has slope \displaystyle -\frac{3}{4}. Through (3, 4):

y - 4 = -\frac{3}{4}(x - 3) \quad\Longrightarrow\quad 3x + 4y = 25 .

To confirm, put \displaystyle y = \frac{25 - 3x}{4} into the equation of the circle and multiply by 16:

\begin{aligned} 16x^2 + (25 - 3x)^2 &= 400 \\ 25x^2 - 150x + 225 &= 0 \\ (x - 3)^2 &= 0 . \end{aligned}

The only solution is x = 3, and then y = 4. One common point: the line is tangent, as the theorem promised. The same idea, with calculus doing the slope, is the equation of a tangent line.

Example 5

Two tangents of equal length

The point A is outside a circle with center O. The lines AB and AC are tangent to the circle at B and C. Prove that AB = AC.

Solution

The tangent theorem supplies two right angles, and hypotenuse-leg finishes the proof.

Statements	Reasons
OB \perp AB and OC \perp AC	A tangent is perpendicular to the radius at the point of contact
\angle OBA = \angle OCA = 90^\circ	Definition of perpendicular
OB = OC	Radii of the same circle
OA = OA	The same segment
\triangle OBA \cong \triangle OCA	Hypotenuse-leg, since OA is the hypotenuse of each right triangle
AB = AC	Corresponding sides of congruent triangles

This is the result behind Two tangents from an outside point, and the next example uses it.

Example 6

A circle inside a right triangle

A circle sits inside a right triangle ABC with its right angle at C, CA = 8 and CB = 6, and touches all three sides. Find its radius.

Solution

The hypotenuse is \displaystyle AB = \sqrt{8^2 + 6^2} = 10. Let the circle have center O and radius r, and touch CB, CA and AB at D, E and F.

Figure 5The three radii to the points of contact are perpendicular to the sides they reach.

Each side is tangent to the circle, so OD \perp CB and OE \perp CA. Look at the quadrilateral CDOE. It has right angles at C, D and E, so it is a rectangle, and two neighboring sides OD and OE are both radii, so they are equal. A rectangle with two equal neighboring sides is a square, and so CD = CE = r.

That leaves BD = 6 - r and AE = 8 - r. From the point B there are two tangents, BD and BF, so they are equal by the previous example; likewise AF = AE. The hypotenuse is the sum of its two pieces:

\begin{aligned} AB &= AF + FB \\ &= (8 - r) + (6 - r) = 14 - 2r . \end{aligned}

Since AB = 10, we get 14 - 2r = 10, so r = 2.

Check. The area is \displaystyle \frac{1}{2}(8)(6) = 24. Splitting the triangle into three triangles that share the center O gives area \displaystyle \frac{1}{2} r (8 + 6 + 10) = 12r. Then 12r = 24 gives r = 2 again.

Example 7

A chord that touches an inner circle

Two circles share a center O, with radii 13 and 5. A chord AB of the larger circle touches the smaller one at T. Find AB.

Solution

Figure 6A chord of the large circle that touches the small circle is perpendicular to OT.

AB is tangent to the small circle at T, so OT \perp AB, and \triangle OTA is right-angled at T with hypotenuse OA = 13, because A is on the large circle. The leg OT = 5 is a radius of the small circle. So

AT^2 = 13^2 - 5^2 = 144 \quad\Longrightarrow\quad AT = 12 .

The same holds on the other side: \triangle OTB is right-angled at T, with OB = 13 and OT = 5, so TB = 12 too. (This is the perpendicular from the center bisects the chord at work.) Therefore

AB = AT + TB = 12 + 12 = 24 .

Boundaries

The word “tangent” is doing the work. Here a tangent is a line with exactly one point in common with the circle. Some books start from a different definition, such as the limit of secant lines or “perpendicular to the radius”, and then the perpendicularity is a definition or a different proof, not this one.

The line, not a segment. The argument needs the whole line to meet the circle once. A segment that stops at P can meet the circle only at P without being perpendicular to OP, and then nothing forces a right angle. A slanted segment ending on the circle from outside is an example.

The radius must go to the point of contact. The right angle is between \ell and OP, where P is the point where they touch. The radius to any other point of the circle is not perpendicular to \ell.

It is a fact about circles. On an ellipse the tangent at a point is generally not perpendicular to the line from the center to that point; it is only at the ends of the axes. The proof here leans on every point of the circle being the same distance r from O, and an ellipse has no such center.

In space the statement survives, with more room. A line tangent to a sphere at P is still perpendicular to the radius OP, by the same proof, since the triangles OQP and OQP' are still congruent. But now there is a whole plane of tangent lines through P, not one.

Where this goes next

With the right angle in hand, tangent problems become right-triangle problems. The equal tangent lengths of Example 5 are the subject of Two tangents from an outside point, and the angle of Example 1 is one half of the arc it cuts off.

The second half of the theorem builds tangents. To draw a tangent from an outside point A, find the points B on the circle where \angle OBA is a right angle; the circle on OA as a diameter does it, and the theorem says the line AB is then tangent.

Circles fitted inside polygons, such as the inscribed circle of a triangle, rest on the same two facts used in Example 6. And common tangents of two circles use the theorem once at each point of contact.

Exercises

Right angles at the point of contact

A line is tangent at P to a circle with center O and radius 9. A point X on the line has PX = 12. Find OX.
A circle has center O and radius 7. From a point A outside it, the tangent AB touches the circle at B, and AB = 24. Find OA.
A circle has radius 6. A point A is 10 from the center, and AB is tangent to the circle at B. Find AB.
The line AB is tangent at B to a circle with center O, and \angle OAB = 28^\circ. Find \angle AOB.
The line \ell is tangent at P to a circle with center O. The chord PA has \angle POA = 110^\circ. Find the acute angle between \ell and the chord PA.
Two tangents from a point A touch a circle with center O at B and C, and \angle BAC = 50^\circ. Find \angle BOC.
Two circles share a center O, with radii 17 and 8. A chord of the larger circle touches the smaller one. Find the length of the chord.
From a point A, two tangents of length 15 are drawn to a circle with center O, and OA = 17. Find the radius of the circle.

Is the line a tangent?

A circle has center O and radius 15. The point P is on the circle, and \ell is a line through P containing a point X with PX = 20 and OX = 25. Is \ell tangent to the circle at P? Give the reason.
A circle has center O and radius 6. The point P is on the circle, and a line through P contains a point X with PX = 7 and OX = 9. Is the line tangent at P? Give the reason.
Is the line 4x + 3y = 20 tangent to the circle x^2 + y^2 = 16? If so, where does it touch?
Is the line x + y = 4 tangent to the circle x^2 + y^2 = 16? Find every point they share.

Coordinates

The circle x^2 + y^2 = 169 passes through (5, 12). Find the equation of the tangent there.
The circle (x - 2)^2 + (y + 1)^2 = 25 passes through (5, 3). Find the equation of the tangent there.
For which values of k is the horizontal line y = k tangent to the circle (x - 1)^2 + (y - 3)^2 = 16?
The tangent to x^2 + y^2 = 100 at (6, 8) crosses the x-axis at a point. Find it.

Reasoning

In the proof that a tangent is perpendicular to the radius, P' is chosen on the other side of Q from P. Explain why P' is different from P, and what happens to the proof if Q were equal to P.
Prove that if \ell is tangent to a circle with center O at P, then every other point of \ell is farther from O than P is.
Two parallel lines are tangent to the same circle, at P and at Q. Prove that PQ is a diameter of the circle.
The proof that a tangent is perpendicular to the radius begins by ruling out the case where the center O is on the line \ell. Why does the rest of the proof not work in that case?
Suppose every point of a line \ell other than P is outside a circle with center O, and P is on the circle and on \ell. Use the fact that the shortest segment from a point to a line is the perpendicular to show that OP \perp \ell.

Answers

OX = 15. The tangent is perpendicular to the radius, so OX^2 = 9^2 + 12^2 = 225.
OA = 25. OB \perp AB, so OA^2 = 7^2 + 24^2 = 625.
AB = 8. OB \perp AB with hypotenuse OA = 10, so AB^2 = 10^2 - 6^2 = 64.
\angle AOB = 62^\circ. The angle at B is 90^\circ, so the other two angles of \triangle OAB sum to 90^\circ, and 90^\circ - 28^\circ = 62^\circ.
55^\circ. \triangle OPA is isosceles, so \displaystyle \angle OPA = \frac{180^\circ - 110^\circ}{2} = 35^\circ. The tangent is at 90^\circ to OP, so the acute angle with the chord is 90^\circ - 35^\circ = 55^\circ. (The other angle between the line and the chord is 125^\circ.)
\angle BOC = 130^\circ. The angles at B and C are right angles, and the four angles of quadrilateral ABOC sum to 360^\circ: 360^\circ - 90^\circ - 90^\circ - 50^\circ = 130^\circ.
The chord is 30. The chord touches the small circle at T, so OT \perp chord and OT = 8. Half the chord is \displaystyle \sqrt{17^2 - 8^2} = 15, and the whole chord is 2 \cdot 15 = 30.
The radius is 8. The radius to a point of contact is perpendicular to the tangent, so r^2 = 17^2 - 15^2 = 64.
Yes. 15^2 + 20^2 = 625 = 25^2, so by the converse of the Pythagorean theorem \angle OPX is a right angle. Then OP \perp \ell, and a line perpendicular to the radius at a point of the circle is tangent there.
No. 6^2 + 7^2 = 85 but 9^2 = 81, so \angle OPX is not a right angle and the line is not perpendicular to OP. A tangent at P must be perpendicular to OP, so this line is not tangent.
Yes, at \displaystyle \left(\frac{16}{5}, \frac{12}{5}\right). Substitute \displaystyle y = \frac{20 - 4x}{3} into the circle: 9x^2 + (20 - 4x)^2 = 144, which simplifies to 25x^2 - 160x + 256 = 0, that is (5x - 16)^2 = 0. One solution, so one common point. (The radius to it has slope \displaystyle \frac{3}{4}, the line has slope \displaystyle -\frac{4}{3}, and the product is -1.)
No. Substitute y = 4 - x: 2x^2 - 8x + 16 = 16, so x(x - 4) = 0. The line meets the circle at (0, 4) and (4, 0). Two points, so it is not tangent.
5x + 12y = 169. The radius to (5, 12) has slope \displaystyle \frac{12}{5}, so the tangent has slope \displaystyle -\frac{5}{12}: \displaystyle y - 12 = -\frac{5}{12}(x - 5), which rearranges to 5x + 12y = 169.
3x + 4y = 27. The center is (2, -1) and the radius to (5, 3) has slope \displaystyle \frac{4}{3}, so the tangent has slope \displaystyle -\frac{3}{4}: \displaystyle y - 3 = -\frac{3}{4}(x - 5), which rearranges to 3x + 4y = 27.
k = 7 or k = -1. A horizontal tangent is perpendicular to a vertical radius, so it touches at the top or bottom of the circle, directly above or below the center (1, 3), at (1, 7) or (1, -1). (Check: substituting y = k gives (x - 1)^2 = 16 - (k - 3)^2, which has one solution only when (k - 3)^2 = 16.)
\displaystyle \left(\frac{50}{3}, 0\right). The radius to (6, 8) has slope \displaystyle \frac{4}{3}, so the tangent has slope \displaystyle -\frac{3}{4} and equation 6x + 8y = 100. Setting y = 0 gives \displaystyle x = \frac{100}{6} = \frac{50}{3}.
Q \neq P because OQ \perp \ell and OP is not, so QP > 0. The point P' is QP away from Q on the other side, so it is not P. If Q were equal to P, then QP = 0 and P' = P: there would be no second point. That is the case OP \perp \ell, where nothing needs to be contradicted.
Let X \neq P be on \ell. By the theorem OP \perp \ell, so \triangle OPX has a right angle at P, and OX^2 = OP^2 + PX^2. Since PX > 0, OX^2 > OP^2, so OX > OP = r.
The radii OP and OQ are perpendicular to their tangents. The two tangents are parallel, so OP and OQ are both perpendicular to the same direction, and there is only one line through O in that direction. So P, O and Q lie on one line. The two tangents are different parallel lines with no point in common, so P \neq Q. A line through O meets the circle at two points, one on each side of O at distance r, so P and Q are those two points and O is the midpoint of PQ. A chord through the center is a diameter.
If O is on \ell, the foot of the perpendicular from O to \ell is O itself, so Q = O and OQ = 0. Then \triangle OQP has no width, and the congruence of \triangle OQP and \triangle OQP' says nothing. That case is settled directly: the line through the center meets the circle at two points, so it is not tangent.
Suppose OP is not perpendicular to \ell, and let Q be the foot of the perpendicular from O. The shortest segment from O to \ell is OQ, and Q \neq P, so OQ < OP. But Q is a point of \ell other than P, so it is outside the circle, which means OQ > r = OP. The two statements contradict each other, so OP \perp \ell.