Chapter 10 · Circles

10.2 Tangents from a Point Outside a Circle

From a point outside a circle you can draw exactly two tangents — and they always have the same length. Prove it, then use it to solve problems on lengths, angles and perimeters, and on quadrilaterals drawn around a circle.

Hold a round tin on a table and put two sharp pencils against it from a point $P$ away from the tin, so that each pencil just touches the tin. You will find two such positions — one on the left of the tin and one on the right. Measure from $P$ to where each pencil touches the tin: the two lengths come out the same. Rotate the picture, change the size of the tin, move $P$ further away — the two lengths are still the same. This is not luck; it is the second big theorem of the chapter.

In this section we prove it, using the right angle at the point of contact from Section 10.1. We then collect its consequences: a formula for the length of a tangent, two angle facts about the figure, and a surprising property of any quadrilateral drawn around a circle. Triangles drawn around a circle (an "incircle") fall out of the same idea.

Two tangents from an outside point

A point $P$ outside a circle lies in a symmetric position with respect to the line $OP$ joining it to the centre: reflect the picture in $OP$ and the circle maps to itself. So there are exactly two tangents from $P$, one on each side of $OP$, touching the circle at $A$ and $B$. (This also agrees with Section 10.1: none from inside, one from a point on the circle, two from outside.)

📘 Definition 10.3 — Length of a tangent
Let $P$ be a point outside a circle and let a tangent from $P$ touch the circle at $A$. The segment $PA$ is called the tangent segment from $P$, and its length $PA$ is the length of the tangent from $P$ to the circle.
📐 Theorem 10.2 — Equal tangents from an external point
The lengths of the two tangents drawn from an external point to a circle are equal. That is, if $PA$ and $PB$ are tangents from $P$ touching the circle with centre $O$ at $A$ and $B$, then $PA=PB$.

Proof. Join $OA$, $OB$ and $OP$.

A second proof, with numbers. In the right-angled triangle $OAP$, Pythagoras gives $PA^2=OP^2-OA^2=OP^2-r^2$, and in $OBP$ it gives $PB^2=OP^2-r^2$. The right-hand sides are equal, so $PA=PB$. This also gives the useful formula $$PA=PB=\sqrt{OP^2-r^2}.$$

What else the congruence tells you. Since $\triangle OAP\cong\triangle OBP$: $\angle APO=\angle BPO$ and $\angle AOP=\angle BOP$. So $PO$ bisects the angle $\angle APB$ between the tangents, and also bisects the angle $\angle AOB$ at the centre. Moreover, in the quadrilateral $OAPB$ the angles at $A$ and $B$ are $90^\circ$ each, and the four angles add up to $360^\circ$, so $$\angle APB+\angle AOB=180^\circ.$$ In words: the angle between the two tangents and the angle that the chord of contact makes at the centre are supplementary. Finally, since $PA=PB$ and $OA=OB$, the line $OP$ is the perpendicular bisector of the chord $AB$.

🧭 Problem-solving strategy — two tangents from a point
  1. Join the centre to $P$ and to both points of contact. Mark the two right angles $\angle OAP=\angle OBP=90^\circ$.
  2. Write $PA=PB$ and note that $OP$ bisects $\angle APB$ and $\angle AOB$.
  3. For lengths: $PA^2=OP^2-r^2$. For angles: use $\angle APB+\angle AOB=180^\circ$, the angle sum of the triangles $OAP$ and $PAB$, and the fact that $PAB$ is isosceles.
  4. Check that the answer is sensible: $PA\gt0$ and $OP\gt r$.
✏️ Example 1 — Length of a tangent
A point $P$ is $17$ cm from the centre $O$ of a circle of radius $8$ cm. Tangents $PA$ and $PB$ are drawn. Find $PA$, $PB$ and the area of the quadrilateral $OAPB$.
Tangent length $\angle OAP=90^\circ$, so $PA^2=OP^2-OA^2=17^2-8^2=289-64=225$ and $PA=15$ cm.
The other tangent By Theorem 10.2, $PB=PA=15$ cm.
Area $OAPB$ is made of two congruent right-angled triangles, each of area $\tfrac12\times8\times15=60$ cm$^2$. Total area $=120$ cm$^2$.
Check $8^2+15^2=64+225=289=17^2$ ✓.
✏️ Example 2 — Angles in the tangent figure
Tangents $PA$ and $PB$ are drawn from $P$ to a circle with centre $O$, and $\angle APB=80^\circ$. Find $\angle POA$ and $\angle AOB$.
Bisector $PO$ bisects $\angle APB$, so $\angle APO=40^\circ$.
Triangle OAP $\angle OAP=90^\circ$, so $\angle POA=180^\circ-90^\circ-40^\circ=50^\circ$.
Angle at the centre $PO$ also bisects $\angle AOB$, so $\angle AOB=2\times50^\circ=100^\circ$.
Check $\angle APB+\angle AOB=80^\circ+100^\circ=180^\circ$ ✓.
✏️ Example 3 — A third tangent and a perimeter
From a point $P$, two tangents $PA$ and $PB$ are drawn to a circle, each of length $12$ cm. A third tangent touches the circle at $C$ (on the arc between $A$ and $B$ nearer to $P$) and meets $PA$ at $X$ and $PB$ at $Y$. Find the perimeter of $\triangle PXY$.
Equal tangents From $X$ the tangents $XA$ and $XC$ are equal, and from $Y$ the tangents $YB$ and $YC$ are equal (Theorem 10.2).
Perimeter $PX+XY+YP=PX+(XC+CY)+YP=(PX+XA)+(YB+YP)=PA+PB$, replacing $XC$ by $XA$ and $CY$ by $YB$.
Value $PA+PB=12+12=24$ cm.
Interactive 1 — Drag the Outside Point: Both Tangents Stay Equal

The circle has radius $r=3$ cm. Drag the red point $P$ with the mouse (or use the two sliders, or press Play). The tangents $PA$ (red) and $PB$ (green) change length as $P$ moves, but they stay equal. The little squares mark the right angles at $A$ and $B$, and the angle $\angle APB$ and the angle $\angle AOB$ always add up to $180^\circ$.

25°6.0
$PA$ and $PB$ (cm)—
$\angle OAP$ and $\angle OBP$—
$\angle APB+\angle AOB$—
$\sqrt{OP^2-r^2}$ (cm)—

Quadrilaterals and triangles drawn around a circle

If every side of a polygon touches a circle, the polygon is said to circumscribe the circle (the circle is then inscribed in the polygon, and is called its incircle). Each vertex is an outside point, and each side is made of two tangent segments. Theorem 10.2 now gives a neat property of four-sided figures.

📐 Theorem 10.3 — Quadrilateral circumscribing a circle
If a quadrilateral $ABCD$ circumscribes a circle (all four sides are tangents to the circle), then the sums of the opposite sides are equal: $$AB+CD=BC+DA.$$

Proof. Let the circle touch $AB$, $BC$, $CD$, $DA$ at $P$, $Q$, $R$, $S$ respectively. The tangents from each vertex are equal (Theorem 10.2):

$AP=AS=a,\quad BP=BQ=b,$
$CQ=CR=c,\quad DR=DS=d.$

Then $AB=AP+PB=a+b$, $BC=BQ+QC=b+c$, $CD=CR+RD=c+d$ and $DA=DS+SA=d+a$. Hence $$AB+CD=(a+b)+(c+d)=(b+c)+(d+a)=BC+DA.\ \blacksquare$$

Consequence. A parallelogram has $AB=CD$ and $BC=DA$, so Theorem 10.3 forces $2AB=2BC$, i.e. $AB=BC$: a parallelogram that circumscribes a circle is a rhombus. The same idea works for a triangle with an incircle: the three vertices give three pairs of equal tangents, and the side lengths determine all six tangent lengths (Example 5).

⚠️ Things to be careful about

1. Label the tangent lengths. Call the two equal tangents from $A$ by the same letter (say $a$), those from $B$ by $b$, and so on. Then each side is the sum of two letters. This avoids sign slips.

2. Opposite, not adjacent. The equal sums are $AB+CD$ and $BC+DA$ — opposite sides. $AB+BC$ is not equal to $CD+DA$ in general.

3. The theorem needs the circle. It is a property of a quadrilateral that circumscribes a circle. If you are told the four side lengths and $AB+CD\ne BC+DA$, then no circle can touch all four sides.

✏️ Example 4 — A quadrilateral with an unknown $x$
A quadrilateral $ABCD$ circumscribes a circle. Its sides are $AB=(3x-2)$ cm, $BC=(x+6)$ cm, $CD=(2x+3)$ cm and $DA=(2x-1)$ cm. Find $x$ and the four sides.
Theorem $AB+CD=BC+DA$ (Theorem 10.3): $(3x-2)+(2x+3)=(x+6)+(2x-1)$.
Solve $5x+1=3x+5$, so $2x=4$ and $x=2$.
Sides $AB=4$ cm, $BC=8$ cm, $CD=7$ cm, $DA=3$ cm (all positive ✓).
Check $AB+CD=4+7=11$ and $BC+DA=8+3=11$ ✓.
✏️ Example 5 — A triangle with an incircle
The incircle of $\triangle ABC$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$. If $AB=12$ cm, $BC=8$ cm and $CA=10$ cm, find $AF$, $BD$ and $CE$.
Label By Theorem 10.2, $AF=AE=x$, $BF=BD=y$, $CD=CE=z$.
Equations $x+y=AB=12$, $y+z=BC=8$, $z+x=CA=10$.
Add $2(x+y+z)=30$, so $x+y+z=15$.
Subtract $x=15-(y+z)=15-8=7$; $y=15-(z+x)=15-10=5$; $z=15-(x+y)=15-12=3$.
Answer $AF=7$ cm, $BD=5$ cm, $CE=3$ cm. Check: $7+5=12$, $5+3=8$, $3+7=10$ ✓.
✏️ Example 6 — The radius of the incircle of a right triangle
In $\triangle ABC$, $\angle B=90^\circ$, $AB=8$ cm and $BC=6$ cm. Find the radius of the circle that touches all three sides.
Hypotenuse $AC=\sqrt{8^2+6^2}=10$ cm.
Figure Let the circle (centre $O$, radius $r$) touch $AB$ at $F$, $BC$ at $D$ and $AC$ at $E$. Then $OF\perp AB$, $OD\perp BC$ and $\angle B=90^\circ$, so $OFBD$ has three right angles and $OF=OD$: it is a square. Hence $BF=BD=r$.
Tangent lengths $AF=AB-BF=8-r$ and $CD=BC-BD=6-r$. By Theorem 10.2, $AE=AF=8-r$ and $CE=CD=6-r$.
Equation $AE+EC=AC$: $(8-r)+(6-r)=10$, so $14-2r=10$ and $r=2$ cm.
Check $\tfrac12\times\text{perimeter}\times r=12\times2=24$ cm$^2$ equals the triangle's area $\tfrac12\times8\times6=24$ cm$^2$ ✓.
Interactive 2 — A Quadrilateral Around a Circle

The circle has radius $2.5$ cm and touches the four sides of $ABCD$ at $P$, $Q$, $R$, $S$. Move the four points of contact with the sliders (or press Play). The two tangents from each vertex have the same colour, because they are equal. Each side is made of two coloured pieces, so $AB+CD$ and $BC+DA$ both equal the sum of all four colours: they always agree.

40°140°215°320°
$AB+CD$ (cm)—
$BC+DA$ (cm)—
tangent lengths from $A$, $B$, $C$, $D$ (cm)—
sides $AB$, $BC$, $CD$, $DA$ (cm)—

Exercises

I
In each part $PA$ and $PB$ are tangents from $P$ to a circle with centre $O$ and radius $r$, touching it at $A$ and $B$.
  • 1 $OP=10$ cm and $r=6$ cm. Find $PA$.
  • 2 $r=5$ cm and $PA=12$ cm. Find $OP$.
  • 3 $OP=17$ cm and $PA=15$ cm. Find $r$.
  • 4 $PA=(2x+3)$ cm and $PB=(4x-5)$ cm. Find $x$ and $PA$.
  • 5 $\angle APB=70^\circ$. Find $\angle AOB$ and $\angle OPA$.
  • 6 $\angle AOB=140^\circ$. Find $\angle APB$, $\angle PAB$ and $\angle OAB$.
  • 7 $\angle PAB=50^\circ$. Find $\angle APB$ and $\angle AOB$.
  • 8 $r=7$ cm and $OP=25$ cm. Find $PA$ and the area of the quadrilateral $OAPB$.
1 $PA=\sqrt{OP^2-r^2}=\sqrt{100-36}=\sqrt{64}=8$ cm.
2 $OP^2=r^2+PA^2=25+144=169$, so $OP=13$ cm.
3 $r^2=OP^2-PA^2=289-225=64$, so $r=8$ cm.
4 $PA=PB$: $2x+3=4x-5$, so $2x=8$ and $x=4$. $PA=2(4)+3=11$ cm (and $PB=4(4)-5=11$ cm ✓).
5 $\angle AOB=180^\circ-70^\circ=110^\circ$. $PO$ bisects $\angle APB$, so $\angle OPA=35^\circ$.
6 $\angle APB=180^\circ-140^\circ=40^\circ$. $PA=PB$, so $\triangle PAB$ is isosceles and $\angle PAB=\tfrac12(180^\circ-40^\circ)=70^\circ$. Also $OA=OB$, so $\angle OAB=\tfrac12(180^\circ-140^\circ)=20^\circ$. (Check: $\angle OAB+\angle PAB=20^\circ+70^\circ=90^\circ=\angle OAP$ ✓.)
7 $PA=PB$, so $\angle PBA=\angle PAB=50^\circ$ and $\angle APB=180^\circ-50^\circ-50^\circ=80^\circ$. Then $\angle AOB=180^\circ-80^\circ=100^\circ$.
8 $PA=\sqrt{625-49}=\sqrt{576}=24$ cm. Area $=2\times\tfrac12\times7\times24=168$ cm$^2$.
II
Third tangents, quadrilaterals and triangles around a circle.
  • 1 Tangents $PA$ and $PB$ ($PA=9$ cm) are drawn from $P$, and a third tangent touches the circle at $C$ and meets $PA$, $PB$ at $X$, $Y$. Find the perimeter of $\triangle PXY$.
  • 2 A quadrilateral $ABCD$ circumscribes a circle, with $AB=9$ cm, $BC=7$ cm and $CD=6$ cm. Find $DA$.
  • 3 A quadrilateral $ABCD$ circumscribes a circle with $AB=(x+3)$, $BC=(2x-1)$, $CD=(x+5)$ and $DA=(2x+1)$ (all in cm). Find $x$ and the four sides.
  • 4 Prove that a parallelogram circumscribing a circle is a rhombus.
  • 5 The incircle of $\triangle ABC$ with $AB=10$ cm, $BC=12$ cm and $CA=14$ cm touches the sides at $D$ ($BC$), $E$ ($CA$), $F$ ($AB$). Find $AF$, $BD$ and $CE$.
  • 6 In $\triangle ABC$, $\angle B=90^\circ$, $AB=15$ cm, $BC=20$ cm. Find the radius of the incircle.
  • 7 In $\triangle ABC$, $\angle B=90^\circ$, $AB=5$ cm, $BC=12$ cm. Find the radius of the incircle.
  • 8 Prove that a rectangle that circumscribes a circle is a square.
1 As in Example 3, the perimeter of $\triangle PXY$ equals $PA+PB=9+9=18$ cm.
2 $AB+CD=BC+DA$: $9+6=7+DA$, so $DA=8$ cm.
3 $(x+3)+(x+5)=(2x-1)+(2x+1)$: $2x+8=4x$, so $x=4$. Sides: $AB=7$, $BC=7$, $CD=9$, $DA=9$ cm; $7+9=16=7+9$ ✓.
4 In parallelogram $ABCD$, $AB=CD$ and $BC=DA$. Theorem 10.3: $AB+CD=BC+DA$, so $2AB=2BC$, i.e. $AB=BC$. All four sides are then equal, so $ABCD$ is a rhombus. $\blacksquare$
5 Let $AF=AE=x$, $BF=BD=y$, $CD=CE=z$. $x+y=10$, $y+z=12$, $z+x=14$. Adding: $x+y+z=18$. $x=18-12=6$, $y=18-14=4$, $z=18-10=8$. So $AF=6$, $BD=4$, $CE=8$ cm (check: $6+4=10$, $4+8=12$, $8+6=14$ ✓).
6 $AC=\sqrt{225+400}=25$ cm. As in Example 6: $(15-r)+(20-r)=25$, so $r=5$ cm.
7 $AC=\sqrt{25+144}=13$ cm. $(5-r)+(12-r)=13$, so $2r=4$ and $r=2$ cm.
8 A rectangle is a parallelogram, so by Part 4 it is a rhombus. A rhombus with a right angle is a square. (Directly: $AB=CD$, $BC=DA$ and $AB+CD=BC+DA$ give $AB=BC$.) $\blacksquare$
III
Proofs and harder calculations (tangents $PA$, $PB$ from $P$ to a circle with centre $O$, as before).
  • 1 Prove that $\angle APB+\angle AOB=180^\circ$.
  • 2 Prove that $OP$ is the perpendicular bisector of $AB$.
  • 3 The angle between the tangents is $60^\circ$ and the radius is $6$ cm. Find $OP$, $PA$ and $AB$ (use $\sqrt3\approx1.732$, answers to 2 decimal places where needed).
  • 4 The tangents are perpendicular to each other ($\angle APB=90^\circ$) and $r=5$ cm. Show that $OAPB$ is a square and find $OP$ to 2 decimal places.
  • 5 $r=5$ cm and $OP=13$ cm. Find the length of the chord $AB$ to 2 decimal places. (Hint: compute the area of $\triangle OAP$ in two ways.)
  • 6 $r=7$ cm, $PA=24$ cm and $OP=25$ cm. Find the chord $AB$.
1 $\angle OAP=\angle OBP=90^\circ$ (Theorem 10.1). The angles of the quadrilateral $OAPB$ add up to $360^\circ$: $\angle APB+\angle AOB=360^\circ-90^\circ-90^\circ=180^\circ$. $\blacksquare$
2 $PA=PB$ (Theorem 10.2) and $OA=OB$ (radii), so both $P$ and $O$ are equidistant from $A$ and $B$. The points equidistant from $A$ and $B$ lie on the perpendicular bisector of $AB$, so the line $OP$ is that perpendicular bisector. $\blacksquare$
3 $PO$ bisects $\angle APB$, so $\angle APO=30^\circ$ and, in right-angled $\triangle OAP$, $\sin30^\circ=\dfrac{OA}{OP}$: $OP=\dfrac{6}{1/2}=12$ cm. $PA=\sqrt{144-36}=\sqrt{108}=6\sqrt3\approx10.39$ cm. In $\triangle PAB$, $PA=PB$ and $\angle APB=60^\circ$, so the other two angles are $60^\circ$ each: it is equilateral and $AB=PA\approx10.39$ cm.
4 $OAPB$ has four right angles ($\angle A=\angle B=90^\circ$ from Theorem 10.1, $\angle P=90^\circ$, so $\angle O=90^\circ$) and $OA=OB$ and $PA=PB$ (so adjacent sides equal): a square of side $5$ cm. $OP$ is its diagonal: $OP=5\sqrt2\approx7.07$ cm.
5 $PA=\sqrt{169-25}=12$ cm. Let $OP$ meet $AB$ at $M$, with $AM\perp OP$. Area of $\triangle OAP=\tfrac12\times OA\times PA=\tfrac12\times5\times12=30$ and also $=\tfrac12\times OP\times AM=\tfrac12\times13\times AM$. So $AM=\dfrac{60}{13}$ and $AB=2AM=\dfrac{120}{13}\approx9.23$ cm.
6 Area of $\triangle OAP=\tfrac12\times7\times24=84=\tfrac12\times25\times AM$, so $AM=\dfrac{168}{25}=6.72$ cm and $AB=13.44$ cm.
IV
Applications.
  • 1 Two straight strings $PA$ and $PB$ touch a circular coin of radius $6$ cm at $A$ and $B$, with $P$ at $10$ cm from the centre of the coin. Find the total length of string $PA+PB$ and the area of the kite $OAPB$.
  • 2 The incircle of $\triangle ABC$ touches $BC$ at $D$ with $BD=6$ cm and $DC=8$ cm. If $CA=12$ cm, find $AB$.
  • 3 In $\triangle ABC$, $\angle B=90^\circ$, $AB=6$ cm and $BC=8$ cm, and $O$ is the centre of the incircle. Find $OA$ to 2 decimal places.
  • 4 Tangents $PA$ and $PB$ with $PA=8$ cm make an angle of $60^\circ$ at $P$. Find $AB$ and the perimeter of $\triangle PAB$.
  • 5 A circular fountain has radius $3.5$ m. Two ropes are stretched from a pole $P$, which is $12.5$ m from the centre of the fountain, to touch the fountain at $A$ and $B$. Find the length of each rope and the total length.
1 $PA=\sqrt{100-36}=8$ cm $=PB$, so $PA+PB=16$ cm. The kite is two right-angled triangles of area $\tfrac12\times6\times8=24$ cm$^2$ each: area $=48$ cm$^2$.
2 Let the incircle touch $CA$ at $E$ and $AB$ at $F$. $CE=CD=8$, so $AE=12-8=4$ cm. $AF=AE=4$ cm and $BF=BD=6$ cm. $AB=AF+FB=4+6=10$ cm.
3 $AC=10$ cm. As in Example 6: $(6-r)+(8-r)=10$, so $r=2$ cm. The tangent from $A$ has length $AF=6-2=4$ cm, and $OF=2$ cm with $\angle OFA=90^\circ$. So $OA=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt5\approx4.47$ cm.
4 $PA=PB$ and $\angle APB=60^\circ$ make $\triangle PAB$ equilateral: $AB=8$ cm and the perimeter is $24$ cm.
5 Each rope: $\sqrt{12.5^2-3.5^2}=\sqrt{156.25-12.25}=\sqrt{144}=12$ m. The total length is $24$ m.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study

How this is examined: the equal-tangents theorem is a standard 3-mark proof, and it is then used in 2–3 mark questions: find an angle in the tangent figure, find a tangent length by Pythagoras, or prove a relation such as $\angle PTQ=2\angle OPQ$. Quadrilaterals around a circle appear as a 3-mark proof or as an equation for $x$. One-mark items are MCQs or Assertion–Reason, and there is often one case study per paper.

E1. Board · 1 mark $PA$ and $PB$ are tangents from $P$ to a circle with centre $O$. If $\angle APB=80^\circ$, then $\angle POA=$  (A) $40^\circ$  (B) $50^\circ$  (C) $80^\circ$  (D) $100^\circ$

Idea $PO$ bisects $\angle APB$, so $\angle APO=40^\circ$. In $\triangle OAP$, $\angle OAP=90^\circ$, so $\angle POA=180^\circ-90^\circ-40^\circ=50^\circ$.
Trap (A) $40^\circ$ is $\angle APO$, a different angle. (D) $100^\circ$ is the full angle $\angle AOB=180^\circ-80^\circ$, twice $\angle POA$.
Answer (B).

E2. Board · 1 mark Assertion (A): A circle can be drawn touching all four sides of a quadrilateral $ABCD$ with $AB=6$ cm, $BC=5$ cm, $CD=4$ cm and $DA=3$ cm. Reason (R): If a quadrilateral $ABCD$ circumscribes a circle, then $AB+CD=BC+DA$.
(A) Both A and R are true and R is the correct explanation of A.  (B) Both A and R are true but R is not the correct explanation of A.  (C) A is true but R is false.  (D) A is false but R is true.

Check R This is Theorem 10.3. True.
Check A $AB+CD=6+4=10$ but $BC+DA=5+3=8$. The sums differ, so by R no circle can touch all four sides: A is false.
Answer (D).

E3. Board · 2 marks The tangent from a point $Q$, which is $29$ cm from the centre $O$ of a circle, has length $21$ cm. Find the radius of the circle.

Right angle If the tangent touches the circle at $T$, then $OT\perp QT$ (Theorem 10.1), so $\triangle OTQ$ is right-angled at $T$ with hypotenuse $OQ=29$ cm. (1 mark)
Radius $OT^2=OQ^2-QT^2=841-441=400$, so the radius is $OT=20$ cm. (1 mark) Check: $20^2+21^2=400+441=841=29^2$ ✓.

E4. Board · 3 marks $TP$ and $TQ$ are the two tangents from an external point $T$ to a circle with centre $O$, touching it at $P$ and $Q$. Prove that $\angle PTQ=2\angle OPQ$.

Isosceles $TP=TQ$ (Theorem 10.2), so in $\triangle TPQ$ the base angles are equal: let $\angle TPQ=\angle TQP=\theta$. Then $\angle PTQ=180^\circ-2\theta$. (1 mark)
Right angle $\angle OPT=90^\circ$ (Theorem 10.1). (1 mark)
Conclude $\angle OPQ=\angle OPT-\angle TPQ=90^\circ-\theta$, so $2\angle OPQ=180^\circ-2\theta=\angle PTQ$. $\blacksquare$ (1 mark)

E5. Board · 3 marks Prove that in a quadrilateral $ABCD$ circumscribing a circle, $AB+CD=BC+DA$.

Setup (1 mark) Let the circle touch $AB$, $BC$, $CD$, $DA$ at $P$, $Q$, $R$, $S$.
Equal tangents (1 mark) Tangents from an external point are equal: $AP=AS$, $BP=BQ$, $CQ=CR$, $DR=DS$.
Add (1 mark) $AB+CD=(AP+BP)+(CR+DR)=(AS+BQ)+(CQ+DS)=(BQ+CQ)+(DS+AS)=BC+DA$. $\blacksquare$

E6. Board · Case study The fountain. A school garden has a circular fountain of radius $6$ m with centre $O$. The gate $G$ is $10$ m from $O$. Two straight paths $GA$ and $GB$ run from the gate and just touch the fountain at $A$ and $B$.
(i) Find the length of the path $GA$, giving the reason for the right angle you use. (2 marks)
(ii) Find the area of the quadrilateral $OAGB$. (1 mark)
(iii) Paving costs $\text{₹}\,250$ per metre of path. Find the cost of paving both paths. (1 mark)

(i) $GA$ is a tangent and $OA$ is the radius to the point of contact, so $OA\perp GA$ (Theorem 10.1). In right-angled $\triangle OAG$: $GA^2=OG^2-OA^2=100-36=64$, so $GA=8$ m. (2 marks)
(ii) $GB=GA=8$ m (Theorem 10.2), and the quadrilateral is two congruent right-angled triangles: area $=2\times\tfrac12\times6\times8=48$ m$^2$.
(iii) Total length $=8+8=16$ m. Cost $=16\times250=\text{₹}\,4000$.
🔬 Research–Teaching NexusMini project · optional, about 30 minutes

Research question. Are the two tangent lengths from an outside point really equal, and does the length follow the formula $\sqrt{OP^2-r^2}$ for a real circular object such as a tin or a bangle?

Why it matters. Whenever a belt, rope or cable passes round a circular drum or pulley, it leaves the drum along a tangent. Engineers must know how much belt is needed from the distance between the pulley centres and the radius, and a rope tied to a point away from a round pillar must be cut to the tangent length. The result also tells a sailor or pilot how far the horizon is: the line of sight to the horizon is a tangent to the Earth.

What to do.

  1. Put a round tin (radius about $3.5$ cm; measure it) on a sheet of paper and trace its outline. Mark the centre $O$ (find it by drawing two chords and their perpendicular bisectors, or by measuring the diameter).
  2. Choose a point $P$ at $6$ cm from $O$. Lay two rulers so that each passes through $P$ and just touches the tin. Mark the two points where the rulers touch the tin as $A$ and $B$.
  3. Measure $PA$ and $PB$, and the angle $\angle OAP$ with a protractor or set-square.
  4. Repeat for $OP=8$ cm and $OP=10$ cm. Make a table of $OP$, $PA$, $PB$ and the predicted length $\sqrt{OP^2-r^2}$.
  5. Write a short explanation of your table using Theorems 10.1 and 10.2.

Tools: A round tin or bangle, plain paper, two rulers, a ruler or tape, a protractor or set-square, a pencil

Hand in: Your sketch, the table with the percentage difference between measured and predicted values, and your explanation.

Going further. Measure $\angle APB$ for each position and check that $\angle APB+\angle AOB=180^\circ$, where $\angle AOB$ is measured at the centre.

Teacher/parent note: for $r=3.5$ cm the predicted tangent lengths are $\sqrt{36-12.25}\approx4.87$ cm at $OP=6$ cm, $\sqrt{64-12.25}\approx7.19$ cm at $OP=8$ cm and $\sqrt{100-12.25}\approx9.37$ cm at $OP=10$ cm. Good measurements agree within about 2 mm. Common errors are measuring $OP$ to the edge of the tin instead of its centre, and a ruler that leans on the tin instead of just touching it.

🧪 Self-check quiz

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Q1. Two tangents $PA$ and $PB$ are drawn from $P$ to a circle. If $PA=7.5$ cm, then $PB=$

Q2. The point $P$ is $13$ cm from the centre of a circle of radius $5$ cm. The length of a tangent from $P$ is

Q3. Tangents $PA$ and $PB$ from $P$ meet at $\angle APB=50^\circ$, and $O$ is the centre. Then $\angle AOB=$

Q4. A quadrilateral $ABCD$ circumscribes a circle, with $AB=10$ cm, $BC=7$ cm and $CD=8$ cm. Then $DA=$

Q5. From a point $P$, tangents $PA$ and $PB$ of length $7$ cm touch a circle. A third tangent touches the circle at $C$ and meets $PA$ at $X$ and $PB$ at $Y$. The perimeter of $\triangle PXY$ is

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