▶ 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
Drag P to move the outside point or O to move the circle; the points of contact follow, and the readout under the canvas keeps PA and PB equal. Learn more about why it holds
Proof?Join the centre O to the points of contact A, B and to P. Each tangent is perpendicular to its radius, so OAP and OBP are right triangles with the same hypotenuse OP and equal legs OA = OB = r. Pythagoras gives PA² = OP² − r² and PB² = OP² − r², so PA = PB. Drag P about the circle and watch the two tangents stay equal.Two tangents from a point
Display
Drag P to move the outside point, O to move the circle
PA = 3.46, PB = 3.46 (OP = 4.00, r = 2.00)
Step 1 / 6
Step explanations
1Two tangents from an outside point
Given P lies outside the circle with centre O. The two lines through P that touch the circle do so at A and at B. To prove: PA = PB.
Step through the proof, then drag the outside point and watch the two tangents swing round the circle while the readout keeps them equal. 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
Proof?Two tangents from a pointDisplayLabelsMeasuresDrag P to move the outside point, O to move the circle
PA = 3.46, PB = 3.46 (OP = 4.00, r = 2.00)
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1Two tangents from an outside point
Given P lies outside the circle with centre O. The two lines through P that touch the circle do so at A and at B. To prove: PA = PB.
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1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names
3Draggable points — P moves the outside point, O the circle; the hollow points of contact A and B follow
4Readout — the line under the figure: PA and PB in the frame's units, with OP and r, live while you drag
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
Proof?Two tangents from a pointDisplayLabelsMeasuresDrag P to move the outside point, O to move the circle
PA = 3.46, PB = 3.46 (OP = 4.00, r = 2.00)
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1Two tangents from an outside point
Given P lies outside the circle with centre O. The two lines through P that touch the circle do so at A and at B. To prove: PA = PB.
2Join O to A, B and P
Construction OA and OB are radii, so OA = OB = r (one tick each). OP is a third segment, shared by the two triangles OAP and OBP that the construction makes.
3Right angles at A and at B
Tangent ⟂ radius A tangent is perpendicular to the radius drawn to its point of contact. So ∠OAP = 90° and ∠OBP = 90°: the triangles OAP and OBP are right triangles with hypotenuse OP.
4PA² = OP² − OA², PB² = OP² − OB²
Pythagorean theorem In the right triangle OAP the hypotenuse is OP, so OP² = OA² + PA², that is PA² = OP² − OA². In the same way, in OBP, PB² = OP² − OB².
5The two right-hand sides are the same
Equal radii OA = OB = r, and OP is the same segment in both triangles. So OP² − OA² and OP² − OB² are the same number, OP² − r², and PA² = PB².
6PA = PB
Positive roots Two lengths with equal squares are equal, since lengths are positive. So PA = PB (two ticks each): the two tangent segments from P are equal. ∎
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1The two tangents — PA and PB, two ticks each at the last step
2The two right triangles — OAP and OBP, sharing the hypotenuse OP, each with a radius for a leg
3The right angles — at the points of contact, by the tangent-radius theorem
5Replay — at the last step Play becomes Replay and Next is off
The Statement
From a point P outside a circle there are exactly two tangents to the circle, touching it at two points A and B. The theorem says that the two tangent segments, from P to the points of contact, are equal:
PA=PB
It is the fact behind "tangent lengths" in every problem about a circle inscribed in a triangle or a quadrilateral, and it is used so often that it is rarely proved. The proof here is short and leans on two earlier pages: the tangent is perpendicular to the radius, and the Pythagorean theorem.
In the player the points of contact A and B are hollow: they are computed from P and the circle, so they swing round as P moves. The readout under the canvas shows PA and PB in the frame's own units, with OP and the radius r beside them, live while you drag.
Opening scene, frozen
P outside the circle, the two tangents touching at A and B.
Joining the Centre
The construction joins the centre O to the three points in sight: to the points of contact, giving the radii OA and OB, and to P, giving the segment OP. The two radii are equal, OA=OB=r, and the player marks them with one tick each.
The three new segments cut the figure into two triangles, OAP and OBP. They share the side OP, and each has a radius for its second side. The proof will show that the third sides, PA and PB, are forced to be equal too.
The construction, frozen
OA, OB, OP joined; the radii marked equal.
Two Right Angles
A tangent is perpendicular to the radius drawn to its point of contact. That was the previous page's theorem, and it applies twice here: OA⊥PA, so the angle at A is right, and OB⊥PB, so the angle at B is right.
The two triangles are therefore right triangles, each with its right angle at the point of contact and with OP as the hypotenuse. Both triangles have the same hypotenuse, since OP is one segment; and both have a leg equal to r. Two right triangles with equal hypotenuses and one pair of equal legs are congruent, and that is in fact the shortest way to finish, but the Pythagorean theorem does the same job with numbers, and this page uses it.
The right angles, frozen
Tangent ⟂ radius at A and at B: two right triangles on the hypotenuse OP.
Pythagoras Twice
In the right triangle OAP the hypotenuse is OP, so the Pythagorean theorem gives OP2=OA2+PA2, that is
PA2=OP2−OA2
In the right triangle OBP the hypotenuse is again OP, so in the same way
PB2=OP2−OB2
The player fills the two triangles in two colours. The two formulas have the same shape; what remains is to see that their right-hand sides are the same number.
Pythagoras twice, frozen
PA² = OP² − OA² in the orange triangle, PB² = OP² − OB² in the blue one.
The Same Numbers
OA and OB are both radii, so OA=OB=r and OA2=OB2=r2. And OP is literally the same segment in both formulas. So
PA2=OP2−r2=PB2
The squares of the two tangent lengths are equal. Notice what the formula says on its own: the square of the tangent length depends only on how far P is from the centre, OP2−r2, a quantity called the power of the point P with respect to the circle.
The same numbers, frozen
OA = OB = r and OP shared: both differences are OP² − r².
Why It Holds
From PA2=PB2 to PA=PB is one step: lengths are positive, and two positive numbers with equal squares are equal. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given tangents, a construction, the tangent-radius theorem, the Pythagorean theorem, equal radii, and the positive square root.
The congruence route gives a little more. Since the right triangles OAP and OBP are congruent, their angles at P are equal, so OP bisects the angle between the two tangents; and their angles at O are equal, so OP also bisects the angle AOB at the centre. Both facts follow from PA=PB as well, by the isosceles base angles theorem applied to the triangle PAB and the symmetry of the figure about OP.
The conclusion, frozen
PA = PB, two ticks each.
Dragging the Figure
Drag P round the circle: the points of contact A and B slide along the circle, the two tangents swing with them, and the readout keeps PA and PB equal to the last digit. Pull P away from the circle and the tangents grow long and nearly parallel; bring it close and they shorten, with A and B closing in on the point of the circle nearest to P. Drag O to move the circle instead.
A drag that would bring P onto or inside the circle is refused, since then there are no tangents from P; so is one that would carry P or O out of the frame.
After dragging P, frozen
P pulled up and closer: the tangents have swung round and shortened, still equal.
Consequences
• Tangent lengths. For a circle inscribed in a triangle, the two tangent segments from each vertex are equal, which gives the lengths s−a, s−b, s−c from the vertices to the points of contact, with s the half perimeter. • Tangential quadrilaterals. A quadrilateral with an inscribed circle has AB+CD=BC+DA, because each side is a sum of two tangent lengths and each tangent length is counted once on each side of the equation. • The angle bisector. The line from an outside point to the centre bisects the angle between the two tangents, and is the perpendicular bisector of the chord of contact AB. • Power of a point.PA2=OP2−r2 does not depend on which tangent is used; the same number equals PX⋅PY for any line through P cutting the circle at X and Y. • Drawing the tangents. The points of contact are where the circle on diameter OP meets the given circle, since the angle at each point of contact is right, as Thales' theorem requires.