Work through in order. Each part changes one thing and keeps the rest still, or keeps the maths still and changes how it looks. Then it asks what you noticed.
This sheet is about
Each circle theorem is a rule about where the points sit, so the first job in any question is to decide which rule the picture is, and check the parts it needs are really there.
Watch for
which rule the picture is · the centre angle is the big one · both angles on the same arc · how the picture is turned
①One problem, no theorem yetStart here
1
Points $A$, $B$ and $C$ lie on a circle with centre $O$. Angle $AOB = 100°$. Tom says angle $ACB$ must be $50°$. Priya says it could be $130°$. Sketch what each of them is picturing and decide who is right.
②Same shape, different ruleSpot the difference
Changes: whether the middle point is the centre or a point on the circleStays the same: the shape of the picture and the $50°$
2
In P, $O$ is the centre. In Q, all four points are on the circle. Tick every true statement (centre rule: the angle at the centre is double; cyclic rule: opposite angles add to $180°$).
Why does the cyclic quadrilateral rule not apply to picture P?
Same kite, two different answers. What do you have to check about the middle point before you choose a rule?
③The centre angle changes, nothing else doesOne thing changes
Changes: the angle at the centreStays the same: the centre $O$, the point $C$ and the rule
3
Angle $AOB$ at the centre is given. Work out $x$ at $C$.
$120°$
=
$150°$
=
$180°$
=
Which of the two angles is always the bigger one? When the centre angle reaches $180°$, what has $AB$ become, and which other circle theorem is that?
④A rule that looks right and is notSpot the difference
Changes: which side of the chord $AB$ the extra point sitsStays the same: the chord $AB$ and the rule each picture looks like
4
In R, $O$ is the centre and angle $ACB = 110°$. In S, angle $ACB = 50°$. Tick every statement that is true.
R
S
R: $x = 220°$R: $x = 55°$R: the angle at $C$ faces the far arc $AB$R: $x = 140°$S: $x = 50°$ (same segment)S: $C$, $D$ on opposite sides of $AB$S: $x = 130°$ (cyclic rule)
R: which angle at $O$ is double $110°$, the marked one or the reflex one? S: what is $x$ if $D$ crosses to the same side as $C$?
In both pictures a rule you know seemed to fit and gave a wrong answer. What had moved, and to where?
⑤The same circle, turned and flippedSame answer, new clothes
Changes: how the picture is turnedStays the same: the three points, the chord $AB$ and the $50°$ at $C$
5
Angle $ACB = 50°$ in each picture. Work out $x$.
upright
=
turned
=
mirrored
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Three pictures, one answer. What did turning the picture change, and what did it leave alone?
⑥Which rule, which arc, which wayAll at once
Changes: the rule, the arc and the direction, all at onceStays the same: the circle and the points named
6
Angle $AOB = 100°$. Work out $x$, and say which angle at $O$ you used.
7
$AB$ is a diameter and angle $CAB = 34°$. Tick every statement that is true.
Angle $ACB = 90°$Angle $CBA = 34°$, angles in the same segment are equalAngle $CBA = 68°$, double the angle at $A$Angle $CBA = 56°$
Which rule, which arc, which direction: all three had to be right at once. Which one did you check first, and why that one?
Teacher page · Circle Theorems
Variation Theory · answers
Object of learning: Each circle theorem is a rule about where the points sit, so the first job in any question is to decide which rule the picture is, and check the parts it needs are really there.
which rule the picture is
the centre angle is the big one
both angles on the same arc
how the picture is turned(something to disregard)
① One problem, no theorem yetStart here
1. Both. With $C$ on the big arc, angle $ACB = 50°$. With $C$ on the small arc between $A$ and $B$, the angle at the centre facing $C$ is the reflex $260°$, so angle $ACB = 130°$.check: $2 \times 50 = 100$ and $2 \times 130 = 260 = 360 - 100$; the two positions of $C$ form a cyclic quadrilateral with $A$ and $B$ and $50 + 130 = 180$. Set before any theorem is named, so "same arc" is met as a problem, not a rule.
② Same shape, different ruleSpot the difference
2. P: $x = 100°$. Q: $x = 130°$. The cyclic rule needs all four vertices on the circle; in P the middle vertex is the centre $O$, so $AOBC$ is not a cyclic quadrilateral.check: P is the ct-centre configuration with $C$ on the major arc, $2 \times 50 = 100$; in Q the points $A$, $D$, $B$, $C$ are in order round the circle so $C$ and $D$ are opposite, $180 - 50 = 130$. Neither rule fits the other picture: $D$ is not the centre, $O$ is not on the circle.
Draw out: Which theorem applies is decided by where the points are, never by the shape: a vertex at the centre means the angle-at-centre rule, four vertices on the circle mean the cyclic quadrilateral rule. The HINGE distractor 80° for a 100° centre angle is the cyclic rule used on the centre figure.
③ The centre angle changes, nothing else doesOne thing changes
$120°$: x = 60°$150°$: x = 75°$180°$: x = 90°
Draw out: The angle at the centre is always the big one, so coming out to the circumference you halve, and going in you double: the SPOT slip is $130°$ at the centre doubled to $260°$, which no triangle can hold. At $180°$ the two radii lie in a straight line, so $AB$ is a diameter and $x = 90°$: the angle in a semicircle is this rule with the centre angle a straight line.
④ A rule that looks right and is notSpot the difference
4. R: $x = 140°$; the reflex angle $AOB$ is the one on the same arc as $C$, $2 \times 110 = 220$, so the marked angle is $360 - 220 = 140$. S: $x = 130°$; with $D$ on the same side as $C$ it would be $50°$.check R: $C$ is on the minor arc, so angle $ACB$ faces the major arc; the centre angle on that arc is the reflex $220°$ and the marked $x$ is the rest of the turn, $140°$; doubling the marked angle gives $220°$, impossible in a triangle, which is the examiner-report "reflex or obtuse" slip (hint $360 - 2x$). Check S: $A$, $C$, $B$, $D$ lie in order round the circle, so $ACBD$ is cyclic and $C$, $D$ are opposite, $180 - 50 = 130$; "same segment" needs the same side of the chord (the HINGE 100° distractor, "thought the two angles equal").
Draw out: The angle-at-centre rule and the same-segment rule both need the two angles to stand on the SAME arc. Move the point across the chord and it looks at the other arc: the centre angle to double is then the reflex one, and the "equal" angle becomes a cyclic-quadrilateral angle, supplementary rather than equal.
⑤ The same circle, turned and flippedSame answer, new clothes
upright: x = 100°turned: x = 100°mirrored: x = 100°
Draw out: Orientation is not part of the theorem, and neither is how big an angle looks: nothing here is drawn accurately, and the examiner note under this topic is students assuming a right angle or an isosceles triangle from appearance. Look for the relation between the points, not where the picture points.
⑥ Which rule, which arc, which wayAll at once
6. $x = 130°$. $C$ is on the small arc, so the angle at $O$ facing $C$ is the reflex angle, $360 - 100 = 260°$, and $x = 260 \div 2 = 130°$.check: $C$ sits between $A$ and $B$ on the minor arc, so angle $ACB$ stands on the major arc; the centre angle on that arc is $260°$; half is $130°$. Also $130 = 180 - 50$, Priya's answer from the opening, now drawn. Three aspects at once: which rule (centre), which arc (the reflex one), which way (halve).
7. Angle $ACB = 90°$ and angle $CBA = 56°$.check: the angle in a semicircle is $90°$ (a centre angle of $180°$ halved, as in part 3), and $180 - 90 - 34 = 56$. The $34°$ and $68°$ are the same-segment rule and the doubling rule used where they do not belong: $A$ and $B$ are not on the same arc from any chord, and neither angle is at the centre.
Draw out: Once each aspect has been met on its own they come together: name the rule from where the points are, find the arc the given angle stands on, then double or halve in the right direction. A student who checks the arc first cannot fall into the reflex trap, and one who names the rule first cannot use the semicircle as a same-segment picture.
Circle Theorems