Maths · 5 Geometry

5.7 Circle theorems II

Before you start

Circle words: centre, radius, diameter, chord, tangent, arc.
The angle between a tangent and the radius at the point of contact is $90^{\circ}$ (5.6).
An isosceles triangle has two equal sides and two equal base angles; angles in a triangle add up to $180^{\circ}$ and angles in a quadrilateral add up to $360^{\circ}$ (5.5).
Pythagoras' theorem, $a^{2} + b^{2} = c^{2}$ , with $c$ the hypotenuse, the side opposite the right angle.
Expanding a bracket such as ${(r - 4)}^{2}$ and solving a simple linear equation.

Syllabus objectives covered (1)

Use the following symmetry properties of circles: • equal chords are equidistant from the centre • the perpendicular bisector of a chord passes through the centre • tangents from an external point are equal in length.

Part 1 of 6

The isosceles triangle inside every chord

A chord is a straight line that joins two points on a circle. Call the ends $A$ and $B$ , and call the centre $O$ .
Join $A$ and $B$ to the centre. $O A$ and $O B$ are both radii, so they are equal. That means triangle $O A B$ is isosceles, with $O$ at the top. Every chord that is not a diameter comes with one of these triangles, whether or not the diagram draws it.
An isosceles triangle has a line of symmetry that goes from the top vertex $O$ to the midpoint $M$ of the base $A B$ , and it meets $A B$ at $90^{\circ}$ . Fold the triangle along $O M$ and $A$ lands exactly on $B$ .
graph diagram — not drawn yet
A circle with centre O and a chord AB below the centre; radii OA and OB drawn and marked equal; a dashed line from O to the midpoint M of AB, with a right-angle mark at M and equal-length marks on AM and MB.
The two radii make triangle $O A B$ isosceles. Its line of symmetry $O M$ halves the chord and meets it at $90^{\circ}$ .
Extend $O M$ in both directions and you get a diameter. A circle looks the same when you reflect it in any diameter, so this line is a line of symmetry of the whole picture, not just of the triangle.
The three facts in this lesson all come from this symmetry. Learn them in the syllabus wording, because that wording is what you write as a reason: equal chords are equidistant from the centre; the perpendicular bisector of a chord passes through the centre; tangents from an external point are equal in length.
The syllabus says you are expected to use these listed properties when you give reasons. So when a question asks for a reason, write the property itself. 'By symmetry' names no property, so it does not tell the reader which fact you used.

Part 2 of 6

The perpendicular bisector of a chord passes through the centre

The perpendicular bisector of $A B$ is the line that does two things at once: it cuts $A B$ exactly in half (bisects it) and it crosses $A B$ at $90^{\circ}$ (is perpendicular to it). A line that does only one of these is not the perpendicular bisector.
Look back at the line $O M$ . It goes through the midpoint of $A B$ and it is perpendicular to $A B$ , so it is the perpendicular bisector of $A B$ . And it starts at $O$ . So the perpendicular bisector of the chord passes through the centre.
There is a second way to see it. The perpendicular bisector of $A B$ is made of exactly the points that are the same distance from $A$ as from $B$ . The centre is the same distance from $A$ and $B$ , because both distances are radii. So the centre must be on that line.
The fact has three forms, and you will meet all of them. A line through $O$ that is perpendicular to a chord bisects the chord. A line from $O$ to the midpoint of a chord is perpendicular to the chord. The perpendicular bisector of a chord passes through $O$ . In each one, two of the three conditions (through $O$ , through the midpoint, at $90^{\circ}$ ) give you the third.
This is what turns a chord into a Pythagoras question. The triangle $O M A$ has a right angle at $M$ . Its hypotenuse is the radius $r$ . One short side is the distance $d$ from the centre, and the other is half the chord.
$r^{2} = d^{2} + {(\frac{1}{2} chord)}^{2}$
For example, in a circle of radius $5$ cm, a chord of $8$ cm has half-length $4$ cm. So $d = \sqrt{5^{2} - 4^{2}} = \sqrt{9} = 3$ cm. Use half the chord, never the whole chord: the right-angled triangle only reaches from $M$ to $A$ .
The same fact finds a centre you cannot see. The centre lies on the perpendicular bisector of every chord. Take two chords that are not parallel. Their perpendicular bisectors cross at exactly one point, and that point lies on both lines, so it is the centre.
graph diagram — not drawn yet
A circle with its centre not marked; two non-parallel chords AB and BC drawn; the perpendicular bisector of each chord drawn as a dashed line with right-angle marks at the chord midpoints; the two dashed lines crossing at a point labelled O.
Each dashed line passes through the centre, so the only point on both of them is the centre.
Why not two parallel chords? Their perpendicular bisectors are both perpendicular to the same direction and both pass through $O$ , so they are the same line. One line does not pin down one point.

Part 3 of 6

Equal chords are equidistant from the centre

Equidistant means 'the same distance from'. The distance from the centre to a chord always means the perpendicular distance: the length $O M$ of the line from $O$ that meets the chord at $90^{\circ}$ . It is the shortest distance from $O$ to the chord. It is not $O A$ , which is just the radius and is the same for every chord.
Now use the right-angled triangle from the last section and make $d$ the subject.
$d = \sqrt{r^{2} - {(\frac{1}{2} chord)}^{2}}$
In one circle, $r$ never changes. Two chords of the same length have the same half-length. Put the same two numbers into the formula and you get the same $d$ . That is the whole reason equal chords are equidistant from the centre.
For example, in a circle of radius $13$ cm, every chord of length $24$ cm has half-length $12$ cm, so it is $\sqrt{13^{2} - 12^{2}} = \sqrt{25} = 5$ cm from the centre, wherever it sits in the circle.
graph diagram — not drawn yet
A circle with centre O and two equal chords AB and CD in different positions, not parallel; perpendiculars OM to AB and ON to CD drawn with right-angle marks, both labelled d; the chords marked equal.
Same radius and same half-chord give the same right-angled triangle, so $O M = O N$ .
The formula works backwards too. Two chords in the same circle that are the same distance from the centre have the same length. So you may also use this fact in the other direction.
The formula tells you one more thing: as $d$ gets bigger, $r^{2} - d^{2}$ gets smaller, so the chord gets shorter. A chord closer to the centre is longer. The longest chord of all is at distance $0$ : it is the diameter.
In a question, the fact saves you a calculation. Once you have worked out the distance of one chord, a chord of the same length in the same circle is the same distance away. Write the distance and the reason: 'equal chords are equidistant from the centre'.

Part 4 of 6

Tangents from an external point are equal in length

A tangent is a straight line that touches a circle at exactly one point, the point of contact. From 5.6 you know that the tangent is perpendicular to the radius at that point.
From a point $P$ outside the circle (an external point) you can draw exactly two tangents. Call their points of contact $A$ and $B$ . The length of a tangent from $P$ means the distance from $P$ to its point of contact: $P A$ or $P B$ .
Draw the radii $O A$ and $O B$ and the line $O P$ . You now have two right-angled triangles, $O A P$ and $O B P$ , with the right angles at $A$ and $B$ because the tangent is perpendicular to the radius.
graph diagram — not drawn yet
A circle with centre O and an external point P to its right; tangents PA and PB touching the circle at A (above) and B (below); radii OA and OB with right-angle marks at A and B; the line OP drawn dashed; PA and PB marked equal and OA and OB marked equal.
Two right-angled triangles with the same hypotenuse $O P$ and the same short side $r$ , so their third sides $P A$ and $P B$ are equal.
Both triangles have hypotenuse $O P$ . Both have a short side equal to the radius $r$ . Pythagoras gives the third side, and it comes out the same in both.
$P A^{2} = O P^{2} - r^{2} = P B^{2}$
So $P A = P B$ : tangents from an external point are equal in length.
For example, if the radius is $5$ cm and $P$ is $13$ cm from the centre, then $P A = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12$ cm, and $P B$ is $12$ cm too.
The trap: the tangent length is not $O P - r$ . In the example, $13 - 5 = 8$ cm is the distance from $P$ to the nearest point of the circle, straight along $O P$ . That point is not $A$ . To get $P A$ you need the right angle at $A$ , so you need Pythagoras.

Part 5 of 6

The kite made by two tangents and two radii

Look at the four-sided shape $O A P B$ . It has $O A = O B$ (radii) and $P A = P B$ (tangents from an external point). Two pairs of equal sides next to each other: it is a kite, and $O P$ is its line of symmetry.
**Angle at the centre plus angle at $P$ .** The angles at $A$ and $B$ are each $90^{\circ}$ , and the four angles of a quadrilateral add up to $360^{\circ}$ . That leaves $180^{\circ}$ for the other two.
$∠ A O B + ∠ A P B = 360^{\circ} - 90^{\circ} - 90^{\circ} = 180^{\circ}$
**The isosceles triangle $P A B$ .** Join $A$ to $B$ . Because $P A = P B$ , triangle $P A B$ is isosceles, so its base angles $P A B$ and $P B A$ are equal. Each is half of what is left after angle $A P B$ .
$∠ P A B = ∠ P B A = \frac{180^{\circ} - ∠ A P B}{2}$
**The second isosceles triangle, $O A B$ .** $O A = O B$ , so angles $O A B$ and $O B A$ are equal too. Angle $O A P$ is $90^{\circ}$ , so angle $O A B = 90^{\circ} - ∠ P A B$ .
** $O P$ halves the angles.** Triangles $O A P$ and $O B P$ have all three sides equal in pairs ( $O A = O B$ , $P A = P B$ , $O P$ shared), so they are identical: each is the reflection of the other in $O P$ . So $O P$ cuts angle $A P B$ in half and cuts angle $A O B$ in half.
** $O P$ is the perpendicular bisector of $A B$ .** $A B$ is a chord. $P$ is the same distance from $A$ and $B$ (equal tangents), so $P$ is on the perpendicular bisector of $A B$ . The perpendicular bisector of a chord passes through the centre, so $O$ is on it too. Two points fix a line, so the line $O P$ is the perpendicular bisector of $A B$ : it meets $A B$ at $90^{\circ}$ at its midpoint.
That last step uses two of this lesson's facts together. It is the kind of chain an Explain question asks for: each sentence states one fact and the property behind it.

Part 6 of 6

Circles inside triangles and quadrilaterals

Sometimes a circle sits inside a triangle and touches all three sides. Then each side is a tangent to the circle, and each vertex is an external point with two tangents coming out of it.
So from each vertex the two tangent lengths are equal. Call the circle's points of contact $P$ on $A B$ , $Q$ on $B C$ and $R$ on $C A$ . Then $A P = A R$ , $B P = B Q$ and $C Q = C R$ .
graph diagram — not drawn yet
A triangle ABC with a circle inside touching AB at P, BC at Q and CA at R; the two tangent lengths from A labelled x, the two from B labelled y, the two from C labelled z.
Each vertex sends out two equal tangent lengths. Each side is made of two different letters.
Give each vertex one letter for its tangent length: $x$ from $A$ , $y$ from $B$ , $z$ from $C$ . Each side is made of two pieces from its two ends.
$A B = x + y, B C = y + z, C A = z + x$
Add all three: the perimeter is $2 (x + y + z)$ , because every letter appears twice. So $x + y + z$ is half the perimeter.
To find one tangent length, take half the perimeter and subtract the side that does not touch that vertex. For $A$ : $x = (x + y + z) - (y + z) = \frac{1}{2} perimeter - B C$ . This works because $B C$ is exactly $y + z$ .
A quadrilateral $A B C D$ whose four sides all touch a circle works the same way. With tangent lengths $a$ , $b$ , $c$ , $d$ from the four vertices, $A B = a + b$ , $B C = b + c$ , $C D = c + d$ and $D A = d + a$ .
So $A B + C D = a + b + c + d$ and $B C + D A = b + c + d + a$ . Both pairs of opposite sides have the same total, because each total uses one tangent length from every vertex.
$A B + C D = B C + D A$

Worked examples

Example 1. A circle has centre $O$ and radius $17$ cm. The chords $A B$ and $C D$ are each $30$ cm long. They are parallel and on opposite sides of $O$ . Calculate the distance between $A B$ and $C D$ . Give reasons.

Show the working

Draw the line from $O$ perpendicular to $A B$ , meeting it at $M$ . The perpendicular bisector of a chord passes through the centre, so this line bisects $A B$ , and $A M = 15$ cm.
Triangle $O M A$ has a right angle at $M$ and hypotenuse $O A = 17$ cm, the radius. Use Pythagoras.
$O M = \sqrt{17^{2} - 15^{2}} = \sqrt{289 - 225} = \sqrt{64} = 8 cm$
$C D$ is the same length as $A B$ , and equal chords are equidistant from the centre. So the perpendicular distance $O N$ from $O$ to $C D$ is also $8$ cm. No second calculation is needed.
The chords are parallel, so the line through $O$ perpendicular to $A B$ is also perpendicular to $C D$ . That means $M$ , $O$ and $N$ lie on one straight line, with $O$ between the chords.
Add the two distances.
$M N = 8 + 8 = 16 cm$

Answer: $16$ cm

Example 2. A horizontal pipe has a circular cross-section. Some water lies in the pipe, filling less than half of it. The surface of the water is $16$ cm wide, and the water is $4$ cm deep at its deepest point. Calculate the radius of the pipe.

Show the working

Model the cross-section as a circle with centre $O$ and radius $r$ . The water surface is a chord $A B$ of length $16$ cm. Let $M$ be its midpoint, so $A M = 8$ cm.
The deepest point of the water is the lowest point of the circle, $C$ , which is directly below the centre. The vertical line through $O$ meets the horizontal chord $A B$ at $90^{\circ}$ , so it is the line through the centre perpendicular to the chord, and it bisects $A B$ . So it passes through $M$ , and $M C = 4$ cm is the depth.
The pipe is less than half full, so the centre is above the water surface. $O C$ is a radius, so $O M = O C - M C = r - 4$ .
Triangle $O M A$ has a right angle at $M$ and hypotenuse $O A = r$ . Write Pythagoras with the unknown in it.
$r^{2} = 8^{2} + {(r - 4)}^{2}$
Expand the bracket: $r^{2} = 64 + r^{2} - 8 r + 16$ . The $r^{2}$ on each side cancels, which is what makes this solvable as a linear equation.
$8 r = 80$
So $r = 10$ . Check: $O M = 10 - 4 = 6$ , and $6^{2} + 8^{2} = 36 + 64 = 100 = 10^{2}$ . Correct.

Answer: The radius is $10$ cm.

Example 3. $P A$ and $P B$ are tangents from the point $P$ to a circle with centre $O$ . They touch the circle at $A$ and $B$ . Angle $A P B = 48^{\circ}$ . Find (a) angle $P A B$ , (b) angle $A O B$ , (c) angle $O A B$ . Give a reason for each answer.

Show the working

(a) Tangents from an external point are equal in length, so $P A = P B$ and triangle $P A B$ is isosceles. Its base angles are equal and share what is left of $180^{\circ}$ .
$∠ P A B = \frac{180^{\circ} - 48^{\circ}}{2} = 66^{\circ}$
(b) The angle between a tangent and a radius is $90^{\circ}$ , so angle $O A P =$ angle $O B P = 90^{\circ}$ . The angles of quadrilateral $O A P B$ add up to $360^{\circ}$ .
$∠ A O B = 360^{\circ} - 90^{\circ} - 90^{\circ} - 48^{\circ} = 132^{\circ}$
(c) Angle $O A P$ is $90^{\circ}$ and is split by $A B$ into angle $O A B$ and angle $P A B$ .
$∠ O A B = 90^{\circ} - 66^{\circ} = 24^{\circ}$
Check (c) another way: $O A = O B$ (radii), so triangle $O A B$ is isosceles and angle $O A B = \frac{180^{\circ} - 132^{\circ}}{2} = 24^{\circ}$ . The two routes agree.

Answer: (a) $66^{\circ}$ ; (b) $132^{\circ}$ ; (c) $24^{\circ}$

Example 4. A circle touches the sides $A B$ , $B C$ and $C A$ of triangle $A B C$ at $P$ , $Q$ and $R$ . $A B = 12$ cm, $B C = 15$ cm and $C A = 11$ cm. Find the lengths $A P$ , $B Q$ and $C R$ .

Show the working

Each side of the triangle is a tangent to the circle. Tangents from an external point are equal in length, so $A P = A R = x$ , $B P = B Q = y$ and $C Q = C R = z$ .
Write each side as its two tangent lengths.
$x + y = 12, y + z = 15, z + x = 11$
Add the three equations. Each letter appears twice, so $2 (x + y + z) = 38$ , and $x + y + z = 19$ .
Subtract each equation from $x + y + z = 19$ to leave one letter: $x = 19 - 15 = 4$ , $y = 19 - 11 = 8$ , $z = 19 - 12 = 7$ .
Check against the sides: $x + y = 4 + 8 = 12$ , $y + z = 8 + 7 = 15$ , $z + x = 7 + 4 = 11$ . All three match.

Answer: $A P = 4$ cm, $B Q = 8$ cm, $C R = 7$ cm

Where people go wrong

Mistake: Putting the whole chord into the right-angled triangle, as in $d = \sqrt{r^{2} - {(chord)}^{2}}$ .

Instead: Use half the chord: $d = \sqrt{r^{2} - {(\frac{1}{2} chord)}^{2}}$ .

The perpendicular from the centre bisects the chord, so the right-angled triangle $O M A$ reaches only from the midpoint $M$ to one end $A$ .

Mistake: Taking the distance from the centre to a chord as the length from $O$ to one end of the chord.

Instead: The distance is the perpendicular distance $O M$ , from $O$ to the chord at $90^{\circ}$ .

$O A$ is a radius, so it is the same for every chord in the circle and tells you nothing about where the chord is.

Mistake: Taking the tangent length $P A$ as $O P - r$ .

Instead: Use the right angle at $A$ : $P A = \sqrt{O P^{2} - r^{2}}$ .

$O P - r$ is the distance from $P$ to the nearest point of the circle along $O P$ , and that point is not the point of contact $A$ . With $r = 5$ and $O P = 13$ , $O P - r = 8$ but $P A = 12$ .

Mistake: Assuming a line passes through the centre because it goes through the midpoint of a chord, or because it is perpendicular to the chord.

Instead: A line through the midpoint must also be perpendicular to the chord (or a perpendicular line must also go through the midpoint) before you can say it passes through the centre.

Many lines pass through the midpoint, and many lines are perpendicular to the chord. Only one line does both, and that is the perpendicular bisector.

Mistake: Writing 'by symmetry' or just 'isosceles' as the reason for equal angles in a tangent diagram.

Instead: Name the property, then the conclusion: 'tangents from an external point are equal in length, so triangle $P A B$ is isosceles'.

The syllabus expects reasons to use the listed geometrical properties. 'Isosceles' says what the triangle is but not why.

Check yourself

A circle has radius $10$ cm. A chord is $12$ cm long. How far is the chord from the centre?
Answer
Half the chord is $6$ cm, so the distance is $\sqrt{10^{2} - 6^{2}} = \sqrt{64} = 8$ cm.
A line from the centre $O$ meets the chord $A B$ at $M$ , at $90^{\circ}$ . $A B = 14$ cm. How long is $A M$ , and why?
Answer
$A M = 7$ cm. The perpendicular bisector of a chord passes through the centre, so the line from $O$ perpendicular to $A B$ bisects it.
Chords $A B$ and $C D$ of the same circle are both $18$ cm long. $A B$ is $5$ cm from the centre. How far is $C D$ from the centre? Calculate the radius of the circle to $3$ significant figures.
Answer
$C D$ is $5$ cm from the centre, because equal chords are equidistant from the centre. Radius $= \sqrt{9^{2} + 5^{2}} = \sqrt{106} = 10.3$ cm ( $3$ s.f.).
The point $P$ is $25$ cm from the centre of a circle of radius $7$ cm. Calculate the length of a tangent from $P$ to the circle.
Answer
$\sqrt{25^{2} - 7^{2}} = \sqrt{625 - 49} = \sqrt{576} = 24$ cm. Both tangents from $P$ are $24$ cm.
$P A$ and $P B$ are tangents to a circle at $A$ and $B$ . Angle $A P B = 70^{\circ}$ . Find angle $P A B$ .
Answer
$P A = P B$ (tangents from an external point are equal in length), so triangle $P A B$ is isosceles: angle $P A B = \frac{180^{\circ} - 70^{\circ}}{2} = 55^{\circ}$ .
Tangents at $A$ and $B$ to a circle with centre $O$ meet at $P$ . Angle $A O B = 110^{\circ}$ . Find angle $A P B$ .
Answer
Angles $O A P$ and $O B P$ are each $90^{\circ}$ , so angle $A P B = 360^{\circ} - 90^{\circ} - 90^{\circ} - 110^{\circ} = 70^{\circ}$ .
A circle touches all three sides of triangle $A B C$ . The tangent lengths from $A$ , $B$ and $C$ are $3$ cm, $5$ cm and $6$ cm. Find the three sides and the perimeter.
Answer
$A B = 3 + 5 = 8$ cm, $B C = 5 + 6 = 11$ cm, $C A = 6 + 3 = 9$ cm. Perimeter $= 28$ cm, which is $2 (3 + 5 + 6)$ .
You are given three points $A$ , $B$ and $C$ on a circle, but not its centre. How can you locate the centre?
Answer
Draw the perpendicular bisectors of the chords $A B$ and $B C$ . The perpendicular bisector of a chord passes through the centre, so the centre is the point where the two lines cross.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. $A$ and $B$ are points on a circle with centre $O$ . The tangents to the circle at $A$ and $B$ meet at the point $P$ . Angle $A O B = 124^{\circ}$ .

(a) Find angle $A P B$ . 2 marks No calculator

Mark scheme

M1 $360 - 90 - 90 - 124$ oe, or $180 - 124$ oe
A1 $56^{\circ}$ cao

(b) Find angle $P A B$ . Give a reason for your answer. 2 marks No calculator

Mark scheme

B1 $62^{\circ}$ ; FT $\frac{180 - their (a)}{2}$
B1 tangents from an external point are equal in length (so triangle $P A B$ is isosceles) oe; or the alternative route: base angles of isosceles triangle $O A B$ (radii) $= 28^{\circ}$ and tangent $⟂$ radius, so $90^{\circ} - 28^{\circ}$

(c) Explain why the line $O P$ is perpendicular to $A B$ . 2 marks No calculator

Mark scheme

B1 $P A = P B$ because tangents from an external point are equal in length, so $P$ lies on the perpendicular bisector of $A B$ oe
B1 the perpendicular bisector of a chord passes through the centre, so $O$ also lies on it and $O P$ is the perpendicular bisector of $A B$ oe. Alternative method: B1 triangles $O A P$ and $O B P$ congruent (e.g. RHS or SSS: $O A = O B$ , $P A = P B$ , $O P$ common), so $O P$ bisects angle $A P B$ ; B1 in isosceles triangle $P A B$ the bisector of the apex angle is perpendicular to the base $A B$

Question 2. A circle has centre $O$ and radius $8.5$ cm. $A B$ is a chord of length $15$ cm and $M$ is the midpoint of $A B$ .

(a) Calculate the length $O M$ . 3 marks

Mark scheme

M1 $A M = 7.5$ seen or used, with angle $O M A = 90^{\circ}$ soi
M1 ${8.5}^{2} - {7.5}^{2}$ oe, or $O M^{2} + {7.5}^{2} = {8.5}^{2}$
A1 $4$ cm cao

(b) $C D$ is another chord of this circle, also of length $15$ cm. State the distance from $O$ to $C D$ . Give a reason for your answer. 2 marks

Mark scheme

B1 $4$ cm; FT their (a)
B1 equal chords are equidistant from the centre oe

(c) $E F$ is a chord of this circle at a distance of $6$ cm from $O$ . Calculate the length of $E F$ . 3 marks

Mark scheme

M1 ${8.5}^{2} - 6^{2}$ $(= 36.25)$ oe
M1 $2 \times \sqrt{their 36.25}$
A1 $12.0$ cm (awrt $12.0$ , e.g. $12.04$ ), or exact $\sqrt{145}$ oe

Question 3. A circle touches the sides $A B$ , $B C$ and $C A$ of triangle $A B C$ at $P$ , $Q$ and $R$ respectively. $A B = 10$ cm, $B C = 13$ cm and $C A = 9$ cm.

(a) Find the length $A P$ . 3 marks No calculator

Mark scheme

M1 equal tangents used to write the sides in tangent lengths, e.g. $x + y = 10$ , $y + z = 13$ , $z + x = 9$ oe
M1 $x + y + z = 16$ , or $\frac{10 + 9 - 13}{2}$ , or another valid elimination oe
A1 $3$ cm cao

Question 4. A circle touches all four sides of the quadrilateral $A B C D$ . It touches $A B$ at $P$ , $B C$ at $Q$ , $C D$ at $R$ and $D A$ at $S$ .

(a) Show that $A B + C D = B C + D A$ . 2 marks No calculator

Mark scheme

B1 $A P = A S$ , $B P = B Q$ , $C Q = C R$ and $D R = D S$ , because tangents from an external point are equal in length
B1 $A B + C D = A P + B P + C R + D R$ and $B C + D A = B Q + C Q + D S + A S$ written, with the equal pairs matched to show the totals are the same

(b) $A B = 9$ cm, $B C = 7$ cm and $C D = 6$ cm. Find the length $D A$ . 1 mark No calculator

Mark scheme

B1 $8$ cm cao

How this is examined

Finding angles in the kite made by two tangents and two radii: a Find or Calculate on a diagram with two tangents from one point, which may also tell you to give a reason for each angle Each angle with the property that gives it: ' $90^{\circ}$ , angle between tangent and radius $= 90^{\circ}$ '; 'base angles equal because tangents from an external point are equal in length'; 'angles in a quadrilateral add up to $360^{\circ}$ '. When reasons are asked for, a correct number without its property does not answer the whole question.
Using the right-angled triangle made by the radius, half the chord and the perpendicular from the centre: a Calculate for the length of a chord, its distance from the centre, or the radius, where the square root may not be exact M1 for half the chord seen (e.g. $7.5$ from a chord of $15$ ); M1 for Pythagoras with the radius as the hypotenuse, e.g. $\sqrt{{8.5}^{2} - {7.5}^{2}}$ ; A1 for the answer, to $3$ significant figures if it is not exact. If the chord is asked for, double the half-length at the end.
Tangent lengths from the corners of a triangle or quadrilateral that has a circle touching every side: a Find for one tangent length or one side, or a Show (that) about the sides The equal pairs written down with the reason (tangents from an external point are equal in length), then each side written as the sum of two tangent lengths. The answer then follows from half the perimeter minus the opposite side.
Stating which circle property gives a length or a distance without working it out: a one- or two-mark State or Write down, followed by 'Give a reason' The value, and the property in the syllabus wording: 'equal chords are equidistant from the centre', 'the perpendicular bisector of a chord passes through the centre', or 'tangents from an external point are equal in length'.
Explaining why a line through an external point and the centre is perpendicular to the chord joining the points of contact: a two-mark Explain with no numbers to calculate Two linked statements: $P A = P B$ (tangents from an external point are equal), so $P$ is on the perpendicular bisector of $A B$ ; the perpendicular bisector of a chord passes through the centre, so $O$ is on it too, so $O P$ is the perpendicular bisector of $A B$ .

Words to know

Chord: A straight line joining two points on a circle. A diameter is the longest chord.
Perpendicular bisector: The line that cuts a line segment in half and crosses it at $90^{\circ}$ . Every point on it is the same distance from both ends of the segment.
Equidistant: The same distance from. For a chord, the distance from the centre is the perpendicular distance.
Tangent: A straight line that touches a circle at exactly one point, the point of contact. It is perpendicular to the radius there.
External point: A point outside the circle. Exactly two tangents can be drawn from it.
Length of a tangent: The distance from the external point to the point of contact, $\sqrt{O P^{2} - r^{2}}$ .
Kite: A quadrilateral with two pairs of equal sides next to each other, such as $O A P B$ formed by two radii and two tangents.

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