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The Tangent Is Perpendicular to the Radius

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag GG to tilt the supposed line ℓ′\ell', TT to move the point of contact, OO to move the circle; the readout under the canvas gives the angle φ\varphi and whether ℓ′\ell' meets the circle once or twice. Learn more about why it holds
  3. Labels toggles the point names, Measures writes the angle φ\varphi beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?Suppose a line through T made an angle other than 90° with the radius OT. Drop the perpendicular from O to that line, meeting it at F, and mark T′ on the other side of F with FT′ = FT. The right triangles OFT and OFT′ are congruent, so OT′ = OT: the line meets the circle a second time, at T′, so it is not a tangent. Drag G towards the tangent and watch T′ slide into T.Tangent ⟂ radius
Display
Drag G to tilt the supposed line, T for the point of contact, O for the centre
OT
φ = 70.3°, OT′ = OT = r: ℓ′ meets the circle twice
Step 1 / 6

Step explanations

1A tangent and the radius to the point of contact
Given
The line ℓ touches the circle with centre O at the single point T: it is a tangent, and OT is the radius to the point of contact. To prove: OT ⟂ ℓ.
No right-angle mark yet: that is what is to be proved. Learn more about the statement

Getting Started with the Visualizer
The Statement
Suppose It Were Not
The Perpendicular and the Mirror Point
Two Congruent Triangles
A Second Common Point
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then drag GG to tilt the supposed line and watch its second crossing point slide into TT only when the angle reaches 90°90°. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
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1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names, Measures writes the angle φ beside its arc
3Draggable points — T moves the point of contact, O the circle; G, which tilts the supposed line, appears at step 2
4Readout — the line under the figure: the angle φ, and whether the supposed line meets the circle once or twice
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 6 of 6
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1The right angle at T — the mark that appears at the last step: the tangent ℓ is perpendicular to OT
2The supposed line ℓ′ — dashed, through T and G, with the second crossing T′ on the circle
3The two right triangles — OFT and OFT′, congruent by SAS, so OT′ = OT = r
4Justification tags — Given, Supposition, Construction, SAS congruence, Contradiction, Conclusion
5Replay — at the last step Play becomes Replay and Next is off

The Statement

A line is a tangent to a circle if it meets the circle at exactly one point, the point of contact. The theorem, Proposition 18 of Book III of Euclid's Elements, says that at the point of contact the tangent is perpendicular to the radius:

OT⊥ℓOT \perp \ell


where OO is the centre, TT the point of contact and ℓ\ell the tangent. It is the fact behind every tangent construction and every "angle between tangent and radius is 90°90°" step in circle problems.

The proof is by contradiction: it shows that any line through TT making an angle other than 90°90° with OTOT meets the circle a second time, so it cannot be the tangent. The player draws such a line, ℓ′\ell', through TT and a draggable point GG, beside the real tangent ℓ\ell, and finds the second crossing. The readout under the canvas shows the angle φ\varphi between OTOT and ℓ′\ell' in degrees, and whether ℓ′\ell' meets the circle once or twice.
OT
Opening scene, frozen

The circle, the tangent ℓ at T, and the radius OT; no right-angle mark yet, that is what is to be proved.


Suppose It Were Not

Suppose a line could touch the circle at TT while making an angle φ≠90°\varphi \neq 90° with the radius OTOT. The dashed line ℓ′\ell' through TT and GG is a line of that kind; the player marks φ\varphi at TT. The claim to be refuted is that ℓ′\ell' meets the circle only at TT.

This is the shape of every proof by contradiction: assume the opposite of what is to be shown, follow it to something impossible, and conclude that the assumption was false. Here the impossible thing will be a second point of ℓ′\ell' on the circle. Nothing is assumed about the size of φ\varphi except that it is not a right angle, which is why you can drag GG anywhere off the line OTOT.
φOTG
The supposition, frozen

The dashed line ℓ′ through T and G, making an angle φ ≠ 90° with OT.


The Perpendicular and the Mirror Point

Drop the perpendicular from OO to ℓ′\ell' and call its foot FF. Since φ≠90°\varphi \neq 90°, the perpendicular from OO is not OTOT itself, so FF is a different point of ℓ′\ell' from TT. In the player FF is hollow: it is built from OO, TT and GG and follows them.

Now mirror TT across FF: on ℓ′\ell', on the far side of FF, mark T′T' with FT′=FTFT' = FT (one tick each). This is the construction of the proof, and it is the same move as in the isosceles base angles proof, where a point was reflected across the foot of the perpendicular; the reflection produces a second triangle that is a copy of the first.
φOTGFT′
The construction, frozen

The perpendicular OF to ℓ′ and the mirror point T′ with FT′ = FT.


Two Congruent Triangles

Look at the triangles OFTOFT and OFT′OFT'. They share the side OFOF; FT=FT′FT = FT' by construction; and the angles at FF, between those two sides, are both right angles, because OFOF is perpendicular to ℓ′\ell'. Two sides and the included angle are equal, so the triangles are congruent by side-angle-side.

Congruent triangles have equal third sides: OT′=OTOT' = OT. The player fills the two triangles in two colours and draws OT′OT'. Everything so far has been honest geometry; the contradiction comes from what OT′=OTOT' = OT means.
φOTGFT′
The congruent triangles, frozen

OFT and OFT′: a common side, equal legs, right angles between; so OT′ = OT.


A Second Common Point

OTOT is a radius, so OT′=OT=rOT' = OT = r says that T′T' is at distance rr from the centre, that is, T′T' lies on the circle (two ticks each). But T′T' also lies on ℓ′\ell', and T′≠TT' \neq T because FF lies strictly between them. So ℓ′\ell' meets the circle at two different points, TT and T′T': it cuts the circle like a chord's line, it does not touch it.

That contradicts the supposition that ℓ′\ell' is a tangent. The only assumption made was φ≠90°\varphi \neq 90°, so that assumption is false.
φOTGFT′
The contradiction, frozen

OT′ = r puts T′ on the circle: ℓ′ meets the circle twice.


Why It Holds

Every line through TT other than the perpendicular to OTOT meets the circle a second time; the tangent ℓ\ell meets it only at TT; so the tangent is the perpendicular, OT⊥ℓOT \perp \ell. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given tangent, a supposition, a construction, side-angle-side congruence, the contradiction, and the conclusion.

A second proof uses the perpendicular is shortest page instead of congruence: if F≠TF \neq T then OF<OT=rOF < OT = r, so FF is inside the circle, and a line through an interior point must leave the circle on both sides, crossing it twice. Both proofs turn on the same picture: the foot of the perpendicular from the centre is the one point of the line closest to OO, and a tangent has exactly one point at distance rr.
φOTGFT′
The conclusion, frozen

The right-angle mark at T: the tangent is the perpendicular to OT.


Dragging the Figure

Drag GG and watch φ\varphi, FF and T′T' change together. The farther φ\varphi is from 90°90°, the farther T′T' sits from TT along ℓ′\ell'. Bring GG towards the tangent ℓ\ell: φ\varphi grows towards 90°90°, FF slides towards TT, and T′T' closes in on TT from the other side; at exactly 90°90° the three points coincide, the readout reports one common point, and ℓ′\ell' has become the tangent. That is the theorem seen the other way round.

Drag TT to move the point of contact (and with it the radius), or OO to move the circle. A drag that puts GG on the line OTOT, or too close to TT, is refused, since then ℓ′\ell' would not be defined.
φOTGFT′
After dragging G, frozen

G pulled up towards the tangent: φ close to 90°, F and T′ crowding in on T.


Consequences

• The converse. The line through TT perpendicular to OTOT is the tangent at TT (Euclid III.16): it is how a tangent is drawn with ruler and compass.
• Distance to a tangent. The distance from the centre to any tangent is exactly rr; a line at distance less than rr cuts the circle, at distance more than rr misses it.
• Two tangents from a point. From a point outside the circle the two tangent segments are equal, since the two right triangles they make with the centre share the hypotenuse and have equal legs rr.
• Tangent and chord. The angle between a tangent and a chord through the point of contact equals the inscribed angle on the other side of the chord.
• Thales' circle again. The tangents from a point PP touch the circle at the two points where the circle on diameter OPOP cuts it, because the angle at each point of contact is right, as Thales' theorem requires.