Geometry

Segment lengths in circles

Two chords cross inside a circle and cut each other into four pieces. Is there a rule tying the four lengths together, and does it survive when the crossing point moves outside the circle?

You need

Chord theoremsGeometry Tangent propertiesGeometry

Unlocks

Proving the power of a point relationshipsGeometry

Where you are 15 in Circles

Geometry11

Circles15

▴ 10 above Angles made by chords, secants and tangents Chord theorems

Segment lengths in circles

Equation of a circle Finding the center and radius

Why this exists

The chord theorems say where a chord sits: how far it is from the center, and that the perpendicular from the center cuts it in half. They say nothing about a chord that is cut somewhere else, by another chord crossing it at a point that is not its midpoint. Draw two chords that cross, and you get four pieces of four different lengths with no obvious relation among them.

There is one, and it is about products, not sums. Multiply the two pieces of one chord, multiply the two pieces of the other, and the answers agree, however the chords are drawn. Even more holds: the same rule works when the point is outside the circle and the lines are secants, and when one line is a tangent. All of it comes from a single number attached to the point, its power, which depends only on how far the point is from the center.

This turns a circle into a tool for finding lengths. Three pieces give the fourth; a tangent’s length follows from one secant; and a circle drawn in coordinates hands over the product without finding a single intersection point.

Definition

A chord is a segment with both ends on the circle. A secant is a line that meets the circle at two points, and a tangent meets it at exactly one; all three are set out in chords, secants and tangents. In the three results below, P is a point that is not on the circle, and PA means the length of \overline{PA}, a distance, so it is never negative.

Two chords crossing inside. Chords \overline{AB} and \overline{CD} cross at P. Then P cuts \overline{AB} into the pieces PA and PB, and cuts \overline{CD} into PC and PD.

Figure 1Two chords crossing at P. The products PA \cdot PB and PC \cdot PD are equal.

Two secants from an outside point. P is outside the circle. One line from P meets the circle first at A and then at B; another meets it first at C and then at D. Here PA is the outside part of the first secant and PB is the whole secant segment, from P to the far point. The chord \overline{AB} is the piece inside the circle, and PB = PA + AB.

Figure 2Two secants from an outside point. Each product is near distance times far distance, PA \cdot PB = PC \cdot PD.

A tangent and a secant. P is outside, the line PT touches the circle at T, and a secant from P meets the circle at A and then B. The tangent segment is \overline{PT}.

Figure 3A tangent and a secant from one point. The tangent’s length squared equals PA \cdot PB.

The power of a point. For a circle with center O and radius r, the power of the point P is the number

OP^2 - r^2

It is negative when P is inside the circle, zero when P is on it, and positive when P is outside.

Intuition

Hold P still and swing a line around it. Each position of the line meets the circle at two points, and the two distances from P change as the line turns. When P is inside, one distance grows exactly as fast, in proportion, as the other shrinks: lengthen one piece by half and the other piece becomes two thirds of what it was. Their product never moves.

The easiest line to measure is the one through the center. If P is inside, that line is a diameter, and the two pieces are r - OP and r + OP. Their product is

(r - OP)(r + OP) = r^2 - OP^2

So if every line through P gives the same product, that is the product, for all of them.

Outside the circle the picture is the same with P beyond both points: the near distance and the far distance trade off so that their product stays fixed, and along the line through the center they are OP - r and OP + r. Now swing the secant until it only grazes the circle. The two meeting points slide together and become the single point T, both distances become PT, and the product becomes PT \cdot PT = PT^2. The tangent is the last secant, with its two points merged.

Results

Intersecting chords. If chords \overline{AB} and \overline{CD} cross at P, then

PA \cdot PB = PC \cdot PD

Two secants. If P is outside a circle, and one line from P meets it at A and then B while another meets it at C and then D, then

PA \cdot PB = PC \cdot PD

In words: outside part times whole, for each secant.

A tangent and a secant. If \overline{PT} is tangent at T and a secant from P meets the circle at A and then B, then

PT^2 = PA \cdot PB

Power of a point. For a circle with center O and radius r, and any line through P that meets the circle at A and B,

PA \cdot PB = \left| OP^2 - r^2 \right|

and when P is outside, every tangent from P has PT^2 = OP^2 - r^2. The three results above are this one statement read for a point inside, a point outside, and a tangent: every product is the same single number.

Why it is true

The whole proof uses two facts you already have: the perpendicular from the center to a chord lands at the chord’s midpoint (chord theorems), and a tangent is perpendicular to the radius at the point of contact (tangent properties). Pythagoras does the rest.

P inside. Take any chord \overline{AB} through P, and drop the perpendicular from O to it, landing at M. Then M is the midpoint of \overline{AB}, so MA = MB; call that length h. Call PM = d, and name the ends so that P is between M and B (if P = M, then d = 0 and nothing below changes).

Figure 4The perpendicular from the center lands at the midpoint M and makes two right triangles, \triangle OMA and \triangle OMP.

Then PA = h + d and PB = h - d, so

PA \cdot PB = (h + d)(h - d) = h^2 - d^2

Now read h^2 and d^2 off the two right triangles with the right angle at M:

\begin{aligned} h^2 &= OA^2 - OM^2 = r^2 - OM^2 \\ d^2 &= OP^2 - OM^2 \end{aligned}

Subtract, and OM^2 cancels:

PA \cdot PB = r^2 - OP^2

The right side never mentions the chord. Every chord through P gives the same product, and that is the intersecting chords result. If the chord passes through O, then M = O, OM = 0, and the same lines hold.

P outside. Draw the same perpendicular to a secant through P, landing at the midpoint M of \overline{AB}. Now P is beyond both points, with A the nearer, so PA = d - h and PB = d + h. The same two right triangles give

PA \cdot PB = d^2 - h^2 = OP^2 - r^2

and again the chord has dropped out, which proves the two secants result.

The tangent. \overline{OT} \perp \overline{PT}, so \triangle OTP has its right angle at T and PT^2 = OP^2 - OT^2 = OP^2 - r^2. That is the same number every secant from P gives, so PT^2 = PA \cdot PB.

There is a second route, through similar triangles: join the ends of the chords so that two triangles appear, find two pairs of equal angles from inscribed angles on the same arc, and read the products off the proportion. It is set out case by case, as formal proofs, in proving the power of a point relationships.

Worked examples

Example 1

Three pieces give the fourth

Chords \overline{AB} and \overline{CD} cross at P, as in Figure 1, with PA = 10, PB = 6 and PC = 5. Find PD and the length of \overline{CD}.

Solution

P is where two chords cross, so the products of the pieces are equal:

\begin{aligned} PA \cdot PB &= PC \cdot PD \\ 10 \cdot 6 &= 5 \cdot PD \\ 60 &= 5 \cdot PD \end{aligned}

Divide both sides by 5: PD = 12.

P lies between C and D, so the whole chord is the sum of its pieces: CD = PC + PD = 5 + 12 = 17.

Check. 5 \cdot 12 = 60 = 10 \cdot 6.

Example 2

Two secants: outside part times whole

From a point P outside a circle, one secant meets the circle at A and then B, with PA = 6 and AB = 10. A second secant meets it at C and then D, with PC = 8. Find PD and CD.

Solution

Both factors in each product are distances from P, so first turn the chord into a distance from P. A is between P and B, so

PB = PA + AB = 6 + 10 = 16

By the two secants result,

\begin{aligned} PC \cdot PD &= PA \cdot PB \\ 8 \cdot PD &= 6 \cdot 16 = 96 \end{aligned}

so PD = 12, and the chord inside the circle is CD = PD - PC = 12 - 8 = 4.

Notice the trade-off: the second secant starts farther out (8 against 6) and so reaches less far (12 against 16). One product fixed means one factor up, the other down.

Example 3

A tangent from the same point

From a point P outside a circle, the tangent segment \overline{PT} has length 12. A secant from P meets the circle at A and then B, with PA = 8. Find PB and AB.

Solution

The tangent and the secant start at the same point, so

\begin{aligned} PT^2 &= PA \cdot PB \\ 144 &= 8 \cdot PB \end{aligned}

and PB = 18. The chord is what is left after the outside part: AB = PB - PA = 18 - 8 = 10.

Check. 8 \cdot 18 = 144 = 12^2.

Example 4

An unknown on both pieces of one chord

Chords \overline{AB} and \overline{CD} cross at P, with PA = x, PB = x + 2, PC = 3 and PD = 8. Find x and the length of each chord.

Solution

Set the products equal:

\begin{aligned} x(x + 2) &= 3 \cdot 8 \\ x^2 + 2x - 24 &= 0 \\ (x + 6)(x - 4) &= 0 \end{aligned}

So x = 4 or x = -6. PA = x is a length and cannot be negative, so x = 4. Then PA = 4 and PB = 6, so AB = 10, and CD = 3 + 8 = 11.

Check. 4 \cdot 6 = 24 = 3 \cdot 8.

Example 5

A tangent and an unknown outside part

From a point P outside a circle, the tangent segment \overline{PT} has length 8. A secant from P cuts off a chord \overline{AB} of length 12 inside the circle, with A nearer to P. Find PA.

Solution

Let PA = x. The whole secant segment is PB = x + 12, and the tangent relation gives

\begin{aligned} x(x + 12) &= 8^2 \\ x^2 + 12x - 64 &= 0 \\ (x + 16)(x - 4) &= 0 \end{aligned}

A length cannot be -16, so PA = 4, and then PB = 16.

Check. 4 \cdot 16 = 64 = 8^2.

Example 6

Every chord through one point

A circle has center O and radius 25, and P is a point inside it with OP = 15.

(a) Find PA \cdot PB for every chord \overline{AB} through P.

(b) Find the length of the chord through P that is perpendicular to \overline{OP}.

Solution

(a) Every chord through P gives the same product, so use the easiest one, the diameter through P. Its pieces are 25 - 15 = 10 and 25 + 15 = 40, and 10 \cdot 40 = 400. The power gives the same: r^2 - OP^2 = 625 - 225 = 400. So PA \cdot PB = 400 for every chord through P.

(b) \overline{OP} is the perpendicular from the center to this chord, so it lands at the chord’s midpoint, and that is P. The two pieces are equal; call each t. Then t^2 = 400, so t = 20 and the chord is 40 long.

The diameter’s pieces, 10 and 40, are the shortest and longest distances from P to the circle, so every piece of every chord through P lies between them. That is why (c) could ask for a piece of 16 but not one of 8. The chord in (b) is the shortest through P: of all pairs of positive numbers with product 400, the pair 20 and 20 has the smallest sum.

Example 7

A circle in coordinates

The circle x^2 + y^2 = 50 has center O = (0, 0), and P = (3, 1). Find PA \cdot PB for the horizontal line through P and for the vertical line through P, and compare both with the power of P.

Solution

Here r^2 = 50 and OP^2 = 3^2 + 1^2 = 10. That is less than 50, so P is inside, and the power is 10 - 50 = -40. The prediction is PA \cdot PB = 40 for every line through P.

The horizontal line y = 1 meets the circle where x^2 + 1 = 50, so x = 7 or x = -7. The points are A = (7, 1) and B = (-7, 1), with PA = 7 - 3 = 4 and PB = 3 - (-7) = 10. The product is 4 \cdot 10 = 40.

The vertical line x = 3 meets the circle where 9 + y^2 = 50, so \displaystyle y = \sqrt{41} or \displaystyle y = -\sqrt{41}. Then \displaystyle PA = \sqrt{41} - 1 and \displaystyle PB = \sqrt{41} + 1, and

(\sqrt{41} - 1)(\sqrt{41} + 1) = 41 - 1 = 40

Both lines give 40, as the power said. The second pair of pieces are irrational, yet their product is a whole number, and the power found it with no square roots at all. For a circle with any center (a, b), the power of P = (x_0, y_0) is (x_0 - a)^2 + (y_0 - b)^2 - r^2: put P into the left side of the equation of the circle and subtract r^2.

Boundaries

The lengths are measured from P. In every product both factors are distances from the point where the lines meet. For a secant from outside, that means the outside part and the whole segment, PA and PB; the chord AB inside the circle is not a factor.

The point must be where the lines meet. The intersecting chords result is about the crossing point of the two chords. A different point on one chord has its own pair of pieces on that chord and says nothing about the other chord.

Inside and outside do not mix. Every line through a point inside the circle meets it twice and cuts out a chord through P, and there is no tangent from P. A point outside has secants and two tangents. A point on the circle has power 0: every line through it has P itself as one meeting point, so one factor is 0 and the products tell you nothing.

The line must reach the circle. A line through an outside point that misses the circle has no A and B, and no product. The tangent is the boundary case, with its two points merged into one.

Only circles. The proof used that every point of the curve is the same distance r from O. On an ellipse the product of the two distances along a line through a fixed point changes as the line turns, so none of these results carries over.

Lengths are positive. When an equation for a length has a negative or zero root, that root is not a length, and only the positive one describes the figure.

Where this goes next

The proof here goes through the center. The other proof, through similar triangles built from inscribed angles, is written out as formal proofs, together with the converse (equal products put four points on one circle), in proving the power of a point relationships.

A chord crossing a diameter at right angles is a special case worth knowing by name. The diameter cuts the chord in half, so the two equal half-chords multiply to the product of the diameter’s two pieces: the half-chord is the geometric mean of those pieces. That is the same fact as the altitude in a right triangle, seen from inside a circle.

This lesson is about the lengths the lines cut off; angles made by chords, secants and tangents is about the angles at the same point, and the two go together in most problems about lines through a circle. And Example 7 points toward coordinates: the power of a point is the left side of the circle’s equation minus r^2, which is how the idea is used in analytic geometry.

Exercises

Two chords crossing inside

In Exercises 1 to 4, chords \overline{AB} and \overline{CD} cross at P.

PA = 9, PB = 4 and PC = 6. Find PD.
PA = 7, PB = 3 and PD = 2.5. Find PC.
PA = 5, PB = 8 and PC = 10. Find PD and CD.
P cuts \overline{AB} into pieces 3 and 12, and it is the midpoint of \overline{CD}. How long is \overline{CD}?

Secants and tangents from outside

In Exercises 5 to 8, P is outside the circle, each secant meets the circle first at the nearer named point, and \overline{PT} is tangent at T.

One secant has PA = 3 and AB = 9. Another has PC = 4. Find PD and CD.
A secant has PA = 5 and PB = 20. Find PT.
PT = 14 and a secant has PA = 7. Find AB.
One secant has PC = 7 and CD = 9. Another has PA = 8. Find AB.

With algebra

Chords cross at P with PA = x, PB = 12, PC = 8 and PD = x + 2. Find x.
Chords cross at P with PA = x, PB = x + 7, PC = 6 and PD = 3x. Find x and all four pieces.
From an outside point, one secant has PA = x and AB = 5, and another has PC = 3 and CD = 9. Find x.
From an outside point, PT = 12, and a secant has PA = x and AB = 2x + 6. Find x.

The power of a point

A circle has radius 15, and P is inside it with OP = 9. Find PA \cdot PB for every chord through P, and the length of the shortest chord through P.
A circle has center O and radius 5, and OP = 13. Find the length of a tangent from P. The line PO meets the circle at A and then B; find PA and PB, and check their product.
A circle has radius 6. A chord through an inside point P is cut by P into pieces whose product is 20. How far is P from the center?
The circle x^2 + y^2 = 65 and the point P = (2, 3) are given. Find the two points where the horizontal line through P meets the circle, then find PA \cdot PB, and compare it with the power of P.
Find the length of a tangent from P = (7, 3) to the circle (x - 2)^2 + (y + 1)^2 = 16.

Reasoning

Two tangents from an outside point P touch a circle at S and T. Use the tangent and secant result to show that PS = PT.
P is on the circle. Explain what PA \cdot PB is for every line through P that meets the circle again, and why the power OP^2 - r^2 agrees.
Chords \overline{AB} and \overline{CD} cross at P with PA = PC. Prove that PB = PD and that the two chords are equally long.
P is outside a circle. Explain why the tangent from P gets longer as P moves farther from the center.
A chord \overline{AB} crosses a diameter \overline{CD} at right angles at P, with PC = a and PD = b. Show that \displaystyle PA = \sqrt{ab}.

Answers

PD = 6. PC \cdot PD = PA \cdot PB = 9 \cdot 4 = 36, so \displaystyle PD = \frac{36}{6}.
PC = 8.4. PC \cdot 2.5 = 7 \cdot 3 = 21, so \displaystyle PC = \frac{21}{2.5}.
PD = 4 and CD = 14. 10 \cdot PD = 5 \cdot 8 = 40, so PD = 4, and CD = 10 + 4.
CD = 12. Each half is t, with t \cdot t = 3 \cdot 12 = 36, so t = 6 and CD = 2t.
PD = 9 and CD = 5. PB = 3 + 9 = 12, so 4 \cdot PD = 3 \cdot 12 = 36, PD = 9, and CD = 9 - 4.
PT = 10. PT^2 = 5 \cdot 20 = 100.
AB = 21. 7 \cdot PB = 14^2 = 196, so PB = 28 and AB = 28 - 7.
AB = 6. PD = 7 + 9 = 16 and 7 \cdot 16 = 112, so \displaystyle PB = \frac{112}{8} = 14 and AB = 14 - 8.
x = 4. 12x = 8(x + 2) = 8x + 16, so 4x = 16. Check: 4 \cdot 12 = 48 = 8 \cdot 6.
x = 11, with pieces PA = 11, PB = 18, PC = 6, PD = 33. x(x + 7) = 18x gives x^2 - 11x = 0, so x(x - 11) = 0, and x = 0 is not a length. Check: 11 \cdot 18 = 198 = 6 \cdot 33.
x = 4. PB = x + 5 and PD = 3 + 9 = 12, so x(x + 5) = 36, x^2 + 5x - 36 = 0, (x + 9)(x - 4) = 0, and the length is the positive root.
x = 6. PB = x + (2x + 6) = 3x + 6, so x(3x + 6) = 144, x^2 + 2x - 48 = 0, (x + 8)(x - 6) = 0. Then PA = 6, PB = 24, and 6 \cdot 24 = 144 = 12^2.
PA \cdot PB = 144, and the shortest chord is 24. r^2 - OP^2 = 225 - 81 = 144. The chord perpendicular to \overline{OP} is cut in half at P, so each half is \displaystyle \sqrt{144} = 12.
PT = 12, PA = 8, PB = 18. PT^2 = 13^2 - 5^2 = 144. Along the line through the center, PA = 13 - 5 and PB = 13 + 5, and 8 \cdot 18 = 144 as it should be.
OP = 4. r^2 - OP^2 = 20, so OP^2 = 36 - 20 = 16.
The points are \displaystyle (2\sqrt{14}, 3) and \displaystyle (-2\sqrt{14}, 3), and PA \cdot PB = 52. On y = 3, x^2 = 65 - 9 = 56, so \displaystyle x = \pm 2\sqrt{14}. The distances from P are \displaystyle 2\sqrt{14} - 2 and \displaystyle 2\sqrt{14} + 2, with product 56 - 4 = 52. The power is 2^2 + 3^2 - 65 = -52, and the product is its absolute value.
PT = 5. The center is (2, -1) and r^2 = 16. OP^2 = 5^2 + 4^2 = 41, so PT^2 = 41 - 16 = 25.
Draw any secant from P, meeting the circle at A and then B (the line through P and the center will do). The tangent and secant result, used twice, gives PS^2 = PA \cdot PB and PT^2 = PA \cdot PB. So PS^2 = PT^2, and since both are positive lengths, PS = PT.
Every such product is 0. Each line through P meets the circle at P itself, so one of the two distances is PP = 0. The power agrees: P on the circle means OP = r, so OP^2 - r^2 = 0.
By the intersecting chords result, PA \cdot PB = PC \cdot PD. Since PA = PC and that length is not 0 (P is not on the circle), divide both sides by it: PB = PD. Then AB = PA + PB = PC + PD = CD.
PT^2 = OP^2 - r^2. The radius r stays the same, so as OP grows, OP^2 - r^2 grows, and so does its positive square root PT.
By the intersecting chords result, PA \cdot PB = PC \cdot PD = ab. The diameter passes through the center and is perpendicular to the chord, so it cuts the chord in half: PA = PB. So PA^2 = ab, and since PA is positive, \displaystyle PA = \sqrt{ab}.