Geometry

Proving the power of a point relationships

Draw any line through a fixed point and multiply the two distances from the point out to where the line meets a circle. The answer never changes. How do you prove that, for a point inside the circle and for a point outside it, and what is the number it keeps giving?

You need

Proving two triangles similarGeometry Segment lengths in circlesGeometry

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

Two chords of a circle cross, and each is cut into two pieces. Measure the pieces and multiply each chord’s pair: the two products come out equal. Move outside the circle and send two lines through it from one point, and the same thing happens with the distances from that point. Replace one of the lines by a tangent, and the square of the tangent’s length joins in. Power of a point states all this as one fact. This lesson proves it.

Reach for it whenever a proof about a circle asks for a length, or for a product or square of lengths, along lines through one point. Reach for its converse when you have lengths and want a circle: equal products put four points on one circle.

Every case is proved the same way, by the pattern of proving two triangles similar: name the correspondence, find two pairs of angles, cite AA, read off a proportion, clear the fractions. What is new here is where the angles come from. The circle supplies them, through inscribed angles that open onto the same arc, so the Inscribed Angle Theorem, proved in proving the inscribed angle theorem, does the real work.

The procedure

The words. A secant is a line that meets a circle at two points. All through this lesson P is a point not on the circle, and every line through P that we use meets the circle. PA is the length of \overline{PA}, a distance, so it is never negative. Two chords cross at P when P is between the ends of each.

Arcs use the notation of the inscribed angle lesson: \overset{\frown}{CB} is an arc from C to B, and an inscribed angle intercepts the arc that lies inside it, the one that does not pass through its vertex. The fact used most below is inscribed angles on the same arc: two inscribed angles that intercept the same arc are congruent, because each measures half of it.

There are three cases, one for each way two lines through P can meet the circle.

Case 1: two chords crossing inside

Given. Chords \overline{AB} and \overline{CD} of one circle lie on different lines and cross at P, so P is between A and B and between C and D.

Prove. PA \cdot PB = PC \cdot PD.

Figure 1Case 1. The angles at A and D both open onto \overset{\frown}{CB}, and the angles at P are vertical angles.

Plan. A product of lengths comes from a proportion by clearing fractions, and a proportion comes from similar triangles. Draw \overline{AC} and \overline{DB} to make \triangle PAC and \triangle PDB. The angles at P are vertical angles. The angle at A and the angle at D both open onto the same arc, \overset{\frown}{CB}. So the angle at A is matched with the angle at D, which fixes the correspondence: P \leftrightarrow P, A \leftrightarrow D, C \leftrightarrow B.

Statements	Reasons
1. P is between A and B, and between C and D	given
2. Draw \overline{AC} and \overline{DB}	auxiliary lines: two points determine a segment
3. {\angle APC \cong \angle DPB}	Vertical Angles Theorem: line 1
4. \angle CAB and \angle CDB both intercept \overset{\frown}{CB}	A and D lie on the same side of line CB, shown below the table
5. {\angle CAB \cong \angle CDB}	inscribed angles on the same arc are congruent: line 4
6. {\angle CAP \cong \angle PDB}	line 5: \angle CAB is \angle CAP because P is on \overrightarrow{AB}, and \angle CDB is \angle PDB because P is on \overrightarrow{DC} (line 1)
7. {\triangle PAC \sim \triangle PDB}	AA similarity: lines 3 and 6
8. \displaystyle {\frac{PA}{PD} = \frac{PC}{PB}}	corresponding sides of similar triangles are proportional: line 7
9. {PA \cdot PB = PC \cdot PD}	multiplication property of equality: multiply line 8 by {PD \cdot PB}

The position fact in line 4. Two inscribed angles are on the same arc only if their vertices are on the same side of the chord, so this needs a reason, and the figure is one drawing. P is between A and B, so \overline{AP} is part of \overline{AB} and does not reach line CB: A is on P’s side of that line. In the same way P is between C and D, so D is on P’s side too. So A and D are on one side of line CB, and the arc from C to B on the other side is the one both angles intercept.

Line 8, read by positions. \overline{PA} is in positions 1 and 2 of \triangle PAC, so its partner is \overline{PD} in positions 1 and 2 of \triangle PDB. \overline{PC} is in positions 1 and 3, so its partner is \overline{PB}.

A second pair of triangles. Drawing \overline{AD} and \overline{CB} instead gives \triangle PAD \sim \triangle PCB, the angles at A and C both intercepting \overset{\frown}{DB}, and the same product again. Exercise 16 writes it out.

Case 2: two secants from an outside point

Given. P is outside the circle. One line through P meets the circle at A and B with A between P and B; a different line through P meets it at C and D with C between P and D.

Prove. PA \cdot PB = PC \cdot PD.

Figure 2Case 2. The angles at B and D both open onto \overset{\frown}{AC}, and both triangles contain the angle at P.

Each product is the outside piece times the whole secant, both measured from P: PB = PA + AB, not AB.

Plan. Draw \overline{AD} and \overline{CB} to make \triangle PAD and \triangle PCB. Both contain the angle at P. The angle at D and the angle at B both open onto \overset{\frown}{AC}, the arc near P. So D is matched with B, and the correspondence is P \leftrightarrow P, A \leftrightarrow C, D \leftrightarrow B.

Statements	Reasons
1. A is between P and B, and C is between P and D	given
2. Draw \overline{AD} and \overline{CB}	auxiliary lines
3. {\angle APD \cong \angle CPB}	reflexive property of congruence: both are the angle between \overrightarrow{PB} and \overrightarrow{PD}, since A is on \overrightarrow{PB} and C on \overrightarrow{PD} (line 1)
4. \angle ADC and \angle ABC both intercept \overset{\frown}{AC}	B and D lie on the same side of line AC, shown below the table
5. {\angle ADC \cong \angle ABC}	inscribed angles on the same arc are congruent: line 4
6. {\angle PDA \cong \angle PBC}	line 5: \angle ADC is \angle PDA because C is on \overrightarrow{DP}, and \angle ABC is \angle PBC because A is on \overrightarrow{BP} (line 1)
7. {\triangle PAD \sim \triangle PCB}	AA similarity: lines 3 and 6
8. \displaystyle {\frac{PA}{PC} = \frac{PD}{PB}}	corresponding sides of similar triangles are proportional: line 7
9. {PA \cdot PB = PC \cdot PD}	multiplication property of equality: multiply line 8 by {PC \cdot PB}

The position fact in line 4. P is not on line AC, since A and C are on different lines through P. A is between P and B, so \overline{PB} crosses line AC at A, and B is on the side of line AC away from P. The same holds for D. So B and D are on one side of line AC and both angles intercept the arc on the other side, the arc nearer P.

A second pair of triangles. Drawing \overline{AC} and \overline{BD} instead gives \triangle PAC \sim \triangle PDB: \angle PAC is the exterior angle at A of the cyclic quadrilateral ABDC, so it is congruent to the interior angle at the opposite corner, D (proving opposite angles of a cyclic quadrilateral, Example 4). Exercise 17 writes it out.

Case 3: a tangent and a secant

Given. P is outside the circle. The line PT is tangent to the circle at T. A second line through P meets the circle at A and B, with A between P and B.

Prove. PT^2 = PA \cdot PB.

Figure 3Case 3. The angle between the tangent and the chord \overline{TA} matches the inscribed angle at B on the same arc.

Plan. Draw \overline{TA} and \overline{TB} to make \triangle PTA and \triangle PBT. Both contain the angle at P. The second pair of angles comes from the tangent chord angle theorem: an angle between a tangent and a chord drawn from the point of contact measures half the arc inside it. So \angle PTA, between the tangent and the chord \overline{TA}, measures half of \overset{\frown}{TA}, and so does the inscribed angle \angle TBA. The angle at T in the first triangle goes with the angle at B in the second, and the correspondence is P \leftrightarrow P, T \leftrightarrow B, A \leftrightarrow T. Here T plays two parts: it is a vertex of both triangles, in position 2 of one and position 3 of the other.

Statements	Reasons
1. A is between P and B; line PT is tangent at T	given
2. Draw \overline{TA} and \overline{TB}	auxiliary lines
3. {\angle TPA \cong \angle BPT}	reflexive property of congruence: A is on \overrightarrow{PB} (line 1), so both are the angle between \overrightarrow{PT} and \overrightarrow{PB}
4. {m\angle PTA = \tfrac{1}{2}\, m\overset{\frown}{TA}}, the arc inside \angle PTA	tangent chord angle: line 1
5. {m\angle TBA = \tfrac{1}{2}\, m\overset{\frown}{TA}}, the same arc	Inscribed Angle Theorem: B is on the other side of line TA from P, shown below the table
6. {\angle PTA \cong \angle TBA}	lines 4 and 5, and the definition of congruent angles
7. {\angle PTA \cong \angle PBT}	line 6: \angle TBA is \angle PBT because A is on \overrightarrow{BP} (line 1)
8. {\triangle PTA \sim \triangle PBT}	AA similarity: lines 3 and 7
9. \displaystyle {\frac{PT}{PB} = \frac{PA}{PT}}	corresponding sides of similar triangles are proportional: line 8
10. {PT^2 = PA \cdot PB}	multiplication property of equality: multiply line 9 by {PB \cdot PT}

The position fact in line 5. Line TA is not line PB, since T is not on line PB (the tangent meets the circle only at T, and line PB is a different line through P). A is between P and B, so P and B are on opposite sides of line TA. The arc inside \angle PTA is on P’s side of \overline{TA}, and B is on the other side, so the arc that \angle TBA intercepts, the one not passing through B, is that same arc.

Line 9, read by positions. \overline{PT} is in positions 1 and 2 of \triangle PTA, and its partner in \triangle PBT is \overline{PB}. \overline{PA} is in positions 1 and 3, and its partner is \overline{PT}. So \overline{PT} appears twice in the proportion, once on top and once below, which is why the result has a square in it: PT is the geometric mean of PA and PB.

The tangent is Case 2 with its two points run together. Turn the lower line of Figure 2 about P, downward and away from the center, and C and D slide toward each other until they meet at a point of tangency T, so PC \cdot PD becomes PT \cdot PT. That is a picture, not a proof; the proof is the table.

The common value

The three cases say that for one point P and one circle, every line through P gives the same product. There is a second proof of this, which also says what the product is. It uses right triangles instead of similar ones.

Given. A circle has center O and radius r. P is not on the circle, and a line through P meets the circle at A and B.

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

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

Proof. Let M be the foot of the perpendicular from O to the line, or M = O if the line passes through O. The perpendicular from the center bisects the chord, so M is the midpoint of \overline{AB}; let h = MA = MB. Put a number line on the line AB with M at 0, A at -h and B at h, and let P be at p. Distances on a number line are absolute differences (distance on a number line), so

Now p^2 = PM^2 and h^2 = MA^2, and both come from the Pythagorean theorem. In the right triangle \triangle OMP, and in the right triangle \triangle OMA whose hypotenuse is the radius \overline{OA}:

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

(If M is O or M is P, a triangle flattens, and each line still holds with OM = 0 or PM = 0.) Subtracting, the OM^2 cancels:

p^2 - h^2 = OP^2 - r^2

and so

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

For a tangent the same number appears at once: the tangent is perpendicular to the radius at T (proving the tangent radius theorem), so \triangle OTP has a right angle at T and PT^2 = OP^2 - r^2.

The power of P. The number OP^2 - r^2, with its sign kept, is called the power of P with respect to the circle. It depends only on how far P is from the center, never on the line.

Where P is	Sign of {OP^2 - r^2}	{PA \cdot PB}
inside, {OP < r}	negative	{r^2 - OP^2}
on the circle, {OP = r}	zero	0
outside, {OP > r}	positive	{OP^2 - r^2}, which is also PT^2

This proof covers all three cases at once, so it is a second, independent proof of each of them. The similar triangle proofs are still worth having: they need no center and no radius, and the similarities themselves are used in other proofs.

When P is on the circle. Then every line through P meets the circle at P itself, so one of the two distances is 0 and every product is 0, which is also OP^2 - r^2. The equation is true and says nothing: it cannot be solved for a length. The similar triangle proofs do not apply at all, since a “triangle” with PA = 0 has collapsed to a segment. All three cases need P off the circle.

The converse

The theorem goes from a circle to equal products. The converse goes back.

Given. Lines AB and CD are different lines meeting at a point P that is none of A, B, C, D. Either P is between A and B and between C and D, or P is between neither pair. And PA \cdot PB = PC \cdot PD.

Prove. A, B, C and D lie on one circle.

The tool is the converse proved in proving opposite angles of a cyclic quadrilateral: a convex quadrilateral with one pair of opposite angles supplementary is cyclic. The equal products give similar triangles by SAS similarity, and the similar triangles give the angles.

P between neither pair. Name the points so that A is between P and B and C is between P and D. Dividing the given product by PD \cdot PB gives \displaystyle \frac{PA}{PD} = \frac{PC}{PB}. \triangle PAC and \triangle PDB share the angle at P, and it is included between the sides in those ratios, so \triangle PAC \sim \triangle PDB by SAS, with A \leftrightarrow D and C \leftrightarrow B. Hence \angle PAC \cong \angle PDB. Now \angle PAC and \angle CAB form a linear pair, and \angle PDB is \angle CDB because C is on \overrightarrow{DP}. So

m\angle CAB + m\angle CDB = 180^\circ

These are opposite angles of quadrilateral ABDC. It is convex: in \triangle PBD, A is on side \overline{PB} and C on side \overline{PD}, so the segments \overline{AD} and \overline{BC}, which are its diagonals, cross inside that triangle. By the converse, ABDC is cyclic.

P between both pairs. Now \overline{AB} and \overline{CD} cross at P, so ACBD is a convex quadrilateral with those two segments as diagonals. The given product gives two proportions, \displaystyle \frac{PA}{PD} = \frac{PC}{PB} and \displaystyle \frac{PA}{PC} = \frac{PD}{PB}, and the angles at P between those sides are vertical angles. So SAS gives \triangle PAC \sim \triangle PDB and \triangle PAD \sim \triangle PCB, and from them \angle PCA \cong \angle PBD and \angle PCB \cong \angle PAD. Add up the angles of ACBD at C and D, each split by the diagonal through P:

\begin{aligned} m\angle ACB &= m\angle PCA + m\angle PCB \\ &= m\angle PBD + m\angle PAD \\ m\angle BDA &= m\angle PDB + m\angle PDA \end{aligned}

Adding, and grouping the angles by triangle, m\angle ACB + m\angle BDA is (m\angle PBD + m\angle PDB) + (m\angle PAD + m\angle PDA). By the triangle angle sum in \triangle PBD and \triangle PAD, the brackets are 180^\circ - m\angle BPD and 180^\circ - m\angle APD, and \angle BPD, \angle APD form a linear pair along line AB, so their measures add to 180^\circ. The total is 360^\circ - 180^\circ = 180^\circ, and ACBD is cyclic by the converse.

The condition on P is needed. If P is between A and B but not between C and D, the four points never lie on one circle, whatever the products are. Exercise 21 asks why.

Worked examples

Example 1

Two chords, and one of them known only as a whole

Chords \overline{AB} and \overline{CD} cross at P. PA = 3, PB = 8 and CD = 10. Find PC and PD.

Solution

Case 1 gives

PC \cdot PD = PA \cdot PB = 3 \cdot 8 = 24

P is between C and D, so by the segment addition postulate PC + PD = CD = 10. Let PC = t; then PD = 10 - t, and

\begin{aligned} t(10 - t) &= 24 \\ t^2 - 10t + 24 &= 0 \\ (t - 4)(t - 6) &= 0 \end{aligned}

So t = 4 or t = 6. Both are possible: the pieces of \overline{CD} are 4 and 6, and nothing in the data says which end is C. Check. 4 \cdot 6 = 24 = 3 \cdot 8, and 4 + 6 = 10.

Example 2

Two secants from one point

From a point P outside a circle, one line meets the circle at A and then B, with PA = 4 and AB = 5. A second line meets it at C and then D, with PC = 3. Find CD.

Solution

Case 2 multiplies the outside piece by the whole secant, both measured from P. The whole of the first secant is

PB = PA + AB = 4 + 5 = 9

PC \cdot PD = PA \cdot PB = 4 \cdot 9 = 36

and \displaystyle PD = \frac{36}{3} = 12. C is between P and D, so

CD = PD - PC = 12 - 3 = 9

Check. The second line has the shorter outside piece, so it must have the longer whole: 12 > 9, and 3 \cdot 12 = 36.

Example 3

A tangent with an unknown outside piece

From P, a tangent touches a circle at T with PT = 6. A secant from P meets the circle at A and then B, with AB = 9. Find PA and PB.

Solution

Let PA = x. Then PB = x + 9, and Case 3 gives

\begin{aligned} x(x + 9) &= 6^2 \\ x^2 + 9x - 36 &= 0 \\ (x + 12)(x - 3) &= 0 \end{aligned}

x = -12 or x = 3. A length is positive, so x = 3: PA = 3 and PB = 12.

Check. 3 \cdot 12 = 36 = 6^2. The tangent, 6, lies between PA and PB, as a geometric mean of two different numbers must.

Example 4

The product from the center

A circle has center O and radius 10, and P is a point with OP = 6. Find the product {PA \cdot PB} for a chord \overline{AB} through P. Then find the pieces of the chord through O, the length of the chord that is perpendicular to \overline{OP} at P, and the other piece of a chord through P that has one piece of length 5.

Solution

OP = 6 < 10, so P is inside, and for every chord through P

PA \cdot PB = r^2 - OP^2 = 100 - 36 = 64

The chord through O. It is a diameter, and P is 6 from the center along it, so its pieces are 10 - 6 = 4 and 10 + 6 = 16. Check. 4 \cdot 16 = 64.

The chord perpendicular to \overline{OP}. \overline{OP} is then the perpendicular from the center, so it bisects the chord and PA = PB. Then PA^2 = 64, PA = 8, and the chord is 16 long. Check with the Pythagorean theorem in \triangle OPA: 6^2 + 8^2 = 100 = 10^2.

A piece of 5. The other piece is \displaystyle \frac{64}{5} = 12.8, and that chord is 5 + 12.8 = 17.8 long.

The three chords have lengths 20, 16 and 17.8. They share one product, and the chord whose pieces are equal is the shortest of them; Exercise 22 shows it is the shortest of all.

Example 5

Two circles and a common tangent

Given. Two circles meet at two points X and Y. A line touches the first circle at S and the second at T. Line XY meets line ST at M.

Prove. M is the midpoint of \overline{ST}.

Figure 5The line through the two crossing points meets a common tangent at M.

Solution

Plan. M lies on a tangent to each circle and on a secant, MXY, that is shared by both. Case 3 in the first circle ties MS to MX \cdot MY, and Case 3 in the second ties MT to the same product.

Statements	Reasons
1. Line ST is tangent to the first circle at S and to the second at T; line XY meets it at M	given
2. M is outside both circles, and X is between M and Y	position facts, below the table
3. MS^2 = MX \cdot MY	Case 3 in the first circle: tangent \overline{MS}, secant through X and Y (lines 1 and 2)
4. MT^2 = MX \cdot MY	Case 3 in the second circle: tangent \overline{MT}, the same secant (lines 1 and 2)
5. MS^2 = MT^2	transitive property of equality: lines 3 and 4
6. MS = MT	both are lengths, so both are positive: take square roots of line 5
7. M is the midpoint of \overline{ST}	M is on line ST and equally far from S and T, and the only such point of the line is the midpoint

The position facts. S \ne T, since two circles tangent to one line at one point touch there and meet nowhere else, and these meet twice. If M were S, then S would be on both circles (it is on line XY and on the first circle, so it is X or Y), and so a second point of the second circle on its tangent line besides T, which is impossible. So M \ne S, and likewise M \ne T. Every point of a tangent line except the point of contact is outside the circle, so M is outside both, hence not between X and Y; naming the nearer crossing point X costs nothing.

The line through the two crossing points of two circles bisects every common tangent. Lengths were never given, and the proof is still about lengths: the products did the work that the numbers do in the earlier examples.

Example 6

Deciding from lengths whether four points are concyclic

(a) Segments \overline{AB} and \overline{CD} cross at P, with PA = 6, PB = 4, PC = 3 and PD = 8. Do A, B, C and D lie on one circle?

(b) Lines AB and CD meet at P, with A between P and B and C between P and D. PA = 2, PB = 9, PC = 3 and PD = 5. Do the four points lie on one circle? If not, where does the circle through A, B and C meet line PD a second time?

Solution

(a) P is between both pairs, and

PA \cdot PB = 6 \cdot 4 = 24 \qquad PC \cdot PD = 3 \cdot 8 = 24

The products are equal, so the converse puts the four points on one circle.

(b) P is between neither pair, and

PA \cdot PB = 2 \cdot 9 = 18 \qquad PC \cdot PD = 3 \cdot 5 = 15

If the four points lay on one circle, Case 2 would force these products to be equal. They are not, so the points are not concyclic. This half uses the theorem, not the converse.

A, B, C are not on one line, so there is exactly one circle through them, the circumscribed circle of \triangle ABC. P is on line AB outside the chord \overline{AB}, so P is outside that circle. Line PC is not tangent to it, since then Case 3 would give PC^2 = 18, and PC^2 = 9. So line PC meets the circle again at a point D', beyond C, and Case 2 gives

PC \cdot PD' = 18 \qquad\Longrightarrow\qquad PD' = \frac{18}{3} = 6

D' is 6 from P and D is 5 from P: D sits 1 unit short of the circle, inside it.

Boundaries

P must not be on the circle. There every product is 0 and the triangles collapse. The equation holds and gives nothing.

The distances are measured from P. In Case 2 the product is the outside piece times the whole secant, PA \cdot PB. The chord \overline{AB} itself enters only through PB = PA + AB.

Each case has its own position facts. Case 1 needs P between the ends of both chords, Case 2 needs P outside, with the near point between P and the far one on each line. Line 4 of each proof uses exactly that (line 4 in Cases 1 and 2, line 5 in Case 3), to put two inscribed angles on the same arc. A proof that skips it has assumed from the figure what the figure cannot show.

The converse needs the matching condition on P. Equal products with P inside one segment and outside the other happen, and the four points are then never concyclic. The converse also needs two different lines: four points on one line lie on no circle.

One circle. The products compare lines through P meeting the same circle. For two different circles the products differ unless P has the same power with respect to both, which is exactly what happens on the line through their two crossing points (Example 5 and Exercise 20).

Flat geometry. Every route above uses the triangle angle sum or the Pythagorean theorem, and both rest on the parallel postulate. On a sphere neither holds, and neither does this theorem.

Where this goes next

The three cases are the three stars under power of a point: two chords crossing inside, two secants from an outside point and a secant and a tangent. With the proofs done they can be quoted as reasons, and they turn many circle problems into one quadratic equation, as in Examples 1 and 3.

The tangent case gives two tangents from one point equal length in a line: PS^2 = PA \cdot PB = PT^2. With a diameter and a perpendicular chord it becomes the geometric mean relation of the altitude to the hypotenuse (Exercise 19). The formula OP^2 - r^2 links the geometry to the equation of a circle: for the circle (x - a)^2 + (y - b)^2 = r^2, putting the point’s coordinates into (x - a)^2 + (y - b)^2 - r^2 gives exactly its power, negative inside and positive outside (point inside or outside a circle).

The converse is a tool for putting points on circles, the move that opens many harder proofs about cyclic quadrilaterals, and Example 5 is the first step toward the line of equal power of two circles, which later courses call the radical axis.

Exercises

Finding lengths

In Exercises 1 to 6, chords \overline{AB} and \overline{CD} cross at P (Exercises 1 and 2), or P is outside the circle with A between P and B and C between P and D (Exercises 3 to 6), and in Exercises 5 and 6, \overline{PT} is tangent at T.

PA = 6, PB = 4 and PC = 3. Find PD.
PA = 5, PB = 7.2 and PD = 9. Find PC.
PA = 5, PB = 12 and PC = 6. Find PD and CD.
PA = 4, AB = 14 and PC = 6. Find CD.
PA = 9 and AB = 16. Find PT.
PT = 10 and PA = 5. Find AB.

With algebra

Chords \overline{AB} and \overline{CD} cross at P, with PA = x, PB = x + 4, PC = 4 and PD = 8. Find x.
Chord \overline{AB} is 13 long and crosses chord \overline{CD} at P, with PC = 4 and PD = 9. Find the two pieces of \overline{AB}.
From an outside point P, one secant has PA = x and AB = 7, and another has PC = x + 1 and CD = 3. Find x and all four distances from P.
From an outside point P, the tangent \overline{PT} has length 2x, and a secant has PA = x and AB = 12. Find x and PT.

The power and the center

A circle has radius 17, and P is inside it with OP = 8. Find PA \cdot PB for any chord \overline{AB} through P, and the length of the chord through P perpendicular to \overline{OP}.
A circle has radius 7, and OP = 25. Find the length of a tangent from P. A secant from P has outside piece PA = 20; find PB and the chord AB.
A circle has radius 6. For points P, Q, R with OP = 4, OQ = 6 and OR = 9, find the power of each point, say where each point is, find PA \cdot PB for a chord through P, and find the length of a tangent from R.
A person’s eye is 1.8 m above a calm sea. Model the Earth as a sphere of radius 6371 km and ignore the bending of light by the air. How far away is the horizon, measured along the line of sight to it? (The line of sight to the horizon is tangent to the Earth, and the eye, the horizon point and the center lie in one plane.)

Is it cyclic?

Decide whether A, B, C and D must lie on one circle, and name what you used. (a) A is between P and B, C between P and D, on two different lines, with PA = 4, PB = 15, PC = 5, PD = 12. (b) The same arrangement with PA = 3, PB = 10, PC = 5, PD = 7. (c) Segments \overline{AB} and \overline{CD} cross at P, with PA = 2.5, PB = 8, PC = 4, PD = 5.

Writing proofs

Given. Chords \overline{AB} and \overline{CD} cross at P. Prove. \triangle PAD \sim \triangle PCB, and PA \cdot PB = PC \cdot PD. Write it as a two-column proof.
Given. P is outside a circle; A is between P and B and C is between P and D, all four on the circle, on two different lines. Prove. \triangle PAC \sim \triangle PDB, and PA \cdot PB = PC \cdot PD, using the exterior angle of the cyclic quadrilateral ABDC.
Given. P is outside a circle, \overline{PT} is tangent at T, and two secants from P meet the circle at A, B and at C, D as in Case 2. Prove. PA \cdot PB = PC \cdot PD without similar triangles for the secants, by using Case 3 twice.
Given. \overline{AB} is a diameter, and chord \overline{CD} is perpendicular to it at P. Prove. PC^2 = PA \cdot PB.
Given. Two circles meet at X and Y. P is on line XY with Y between X and P. A line through P meets the first circle at A and B with A between P and B; a different line through P meets the second circle at C and D with C between P and D. Prove. PA \cdot PB = PC \cdot PD, and A, B, C, D lie on one circle.

Reasoning

Lines AB and CD meet at P, with P between A and B but C between P and D, and PA \cdot PB = PC \cdot PD. Explain why A, B, C and D cannot lie on one circle.
P is inside a circle and is not its center. Prove that of all chords through P, the one perpendicular to \overline{OP} is the shortest.
In the proof of Case 1, explain why the similarity is written \triangle PAC \sim \triangle PDB and not \triangle PAC \sim \triangle PBD. What proportion would the second statement give, and why is it false in general?

Answers

PD = 8. PC \cdot PD = PA \cdot PB = 24, so \displaystyle PD = \frac{24}{3}.
PC = 4. PA \cdot PB = 5 \times 7.2 = 36, so \displaystyle PC = \frac{36}{9}.
PD = 10 and CD = 4. PC \cdot PD = 5 \cdot 12 = 60, so \displaystyle PD = \frac{60}{6} = 10, and CD = PD - PC = 10 - 6.
CD = 6. PB = 4 + 14 = 18, so PC \cdot PD = 4 \cdot 18 = 72, PD = 12 and CD = 12 - 6.
PT = 15. PB = 9 + 16 = 25, and PT^2 = 9 \cdot 25 = 225.
AB = 15. \displaystyle PB = \frac{PT^2}{PA} = \frac{100}{5} = 20, and AB = PB - PA = 20 - 5.
x = 4. x(x + 4) = 32 gives x^2 + 4x - 32 = 0, that is (x + 8)(x - 4) = 0, and a length is positive. Check: 4 \cdot 8 = 32 = 4 \cdot 8.
The pieces are 4 and 9. With PA = t and PB = 13 - t, t(13 - t) = 36, so t^2 - 13t + 36 = 0, (t - 4)(t - 9) = 0. Either end may be A.
x = 2, and PA = 2, PB = 9, PC = 3, PD = 6. PB = x + 7 and PD = x + 4, so x(x + 7) = (x + 1)(x + 4), which is x^2 + 7x = x^2 + 5x + 4, so 2x = 4. Check: 2 \cdot 9 = 18 = 3 \cdot 6.
x = 4 and PT = 8. (2x)^2 = x(x + 12) gives 4x^2 = x^2 + 12x, so 3x^2 = 12x, and since x > 0, x = 4. Check: PA \cdot PB = 4 \cdot 16 = 64 = 8^2.
PA \cdot PB = 225, and the chord is 30 long. r^2 - OP^2 = 289 - 64 = 225. The perpendicular from O bisects that chord at P, so each half is \displaystyle \sqrt{225} = 15.
The tangent is 24 long, PB = 28.8 and AB = 8.8. PT^2 = 25^2 - 7^2 = 576. Then \displaystyle PB = \frac{576}{20} = 28.8 and AB = 28.8 - 20, which is less than the diameter 14, as a chord must be.
Powers -20, 0 and 45: P is inside, Q on the circle, R outside. 4^2 - 36 = -20, 6^2 - 36 = 0, 9^2 - 36 = 45. For a chord through P, PA \cdot PB = 20, and the tangent from R has length \displaystyle \sqrt{45} = 3\sqrt{5} \approx 6.71.
About 4.8 km. With h = 0.0018 km, the eye is 6371 + h km from the center, and the line of sight is tangent, so d^2 = (6371 + h)^2 - 6371^2, which is h(2 \cdot 6371 + h) \approx 22.94, and d \approx 4.79 km. It is also Case 3 with the secant through the center: outside piece h, whole secant h + 12742.
(a) Yes, by the converse: 4 \cdot 15 = 60 = 5 \cdot 12, and P is between neither pair. (b) No, by the theorem: 3 \cdot 10 = 30 but 5 \cdot 7 = 35, and Case 2 would force them equal. (c) Yes, by the converse: 2.5 \cdot 8 = 20 = 4 \cdot 5, and P is between both pairs.
(1) P is between A and B and between C and D (given). (2) Draw \overline{AD} and \overline{CB} (auxiliary lines). (3) \angle APD \cong \angle CPB (Vertical Angles Theorem: line 1). (4) \angle DAB \cong \angle DCB (inscribed angles on the same arc \overset{\frown}{DB}; A and C are both on P’s side of line DB because P is between A and B and between C and D). (5) \angle PAD \cong \angle PCB (line 4, since P is on \overrightarrow{AB} and on \overrightarrow{CD}). (6) \triangle PAD \sim \triangle PCB (AA similarity: lines 3 and 5). (7) \displaystyle \frac{PA}{PC} = \frac{PD}{PB} (corresponding sides: line 6). (8) PA \cdot PB = PC \cdot PD (multiplication property: line 7).
(1) A, B, D, C lie on the circle in that order, since \overline{AD} and \overline{BC} cross inside \triangle PBD (given positions). (2) \angle PAC \cong \angle BDC (each exterior angle of a cyclic quadrilateral is congruent to the interior angle at the opposite corner: the exterior angle at A of ABDC, since A is between P and B). (3) \angle PDB is \angle CDB (C is on \overrightarrow{DP}), so \angle PAC \cong \angle PDB. (4) \angle APC \cong \angle DPB (reflexive: both are the angle at P between the two lines). (5) \triangle PAC \sim \triangle PDB (AA: lines 3 and 4). (6) \displaystyle \frac{PA}{PD} = \frac{PC}{PB} (corresponding sides: line 5). (7) PA \cdot PB = PC \cdot PD (multiplication property: line 6).
(1) PT^2 = PA \cdot PB (Case 3 with the tangent and the first secant). (2) PT^2 = PC \cdot PD (Case 3 with the tangent and the second secant). (3) PA \cdot PB = PC \cdot PD (transitive and symmetric properties: lines 1 and 2).
(1) \overline{AB} is a diameter and \overline{CD} \perp \overline{AB} at P (given). (2) PC = PD (a diameter perpendicular to a chord bisects it: line 1). (3) PA \cdot PB = PC \cdot PD (Case 1: the chords cross at P, which is between C and D by line 2 and inside the circle, so between A and B). (4) PA \cdot PB = PC^2 (substitution: line 2 into line 3), and by the symmetric property PC^2 = PA \cdot PB.
(1) P is outside both circles and Y is between P and X (given: P is on line XY outside the chord \overline{XY} of each circle). (2) PA \cdot PB = PY \cdot PX (Case 2 in the first circle). (3) PC \cdot PD = PY \cdot PX (Case 2 in the second circle). (4) PA \cdot PB = PC \cdot PD (transitive and symmetric properties: lines 2 and 3). (5) A, B, C, D lie on one circle (the converse: two different lines through P, and P is between neither pair). If line PA is line XY itself, A and B are Y and X; the argument does not change.
Suppose they did lie on one circle. P is between A and B, two points of that circle, so P is inside it. A line through a point inside a circle meets the circle once on each side of the point, so P would be between C and D. But C is between P and D, so P is not between them, a contradiction. The equal products are a coincidence of the numbers, not a sign of a circle.
Every chord \overline{AB} through P has PA \cdot PB = r^2 - OP^2, one fixed positive number k. Its length is PA + PB, and (PA + PB)^2 equals (PA - PB)^2 + 4\,PA \cdot PB, which is (PA - PB)^2 + 4k. This is least exactly when PA = PB, that is when P is the midpoint of the chord. The chord with midpoint P is the one perpendicular to \overline{OP}, since the segment from the center to a chord’s midpoint is perpendicular to the chord, and there is only one line through P perpendicular to \overline{OP}.
The angles fix the correspondence. The pair of angles proved congruent is the angle at A and the angle at D (both on \overset{\frown}{CB}), so A goes with D, and then C goes with B. \triangle PAC \sim \triangle PBD would claim \angle PAC \cong \angle PBD, which nothing proves, and would give \displaystyle \frac{PA}{PB} = \frac{PC}{PD}, that is PA \cdot PD = PB \cdot PC. In Exercise 1 that reads 6 \cdot 8 = 4 \cdot 3, which is false.