Solving exam questions · Circle geometry

Objective questions usually need one theorem. Theory questions chain two or three together, and the hard part is seeing which one to use first. This lesson gives you a routine for that.

A routine for any circle question

Copy the diagram and mark everything you’re given. Mark equal lengths too, not just angles.
Hunt for the special features. Each one points to a theorem:

If you see…	Think…
The centre O with two radii	Isosceles triangle; angle at the centre
A line through O from one side of the circle to the other	Diameter: angle in a semicircle is $90^\circ$
Two angles on the same chord, same side	Angles in the same segment are equal
Four points on the circle joined up	Cyclic quadrilateral: opposite angles add to $180^\circ$
A side of a cyclic quadrilateral carried on past a corner	Exterior angle = interior opposite angle
A tangent	$90^\circ$ with the radius; alternate segment theorem
Two tangents from one point	Equal lengths

Work from what you know towards what you want. Each new angle should use one fact.
Write the reason beside every angle you find, in brackets.

If you’re stuck, the answer is often an extra line: join two points that aren’t joined yet, usually to the centre or to make a triangle on a diameter.

WAEC 2015: the line nobody drew

This question has two circles. The trick is one extra line. Step through it and watch the diagram: it shows only what each step uses.

Worked example · WAEC 2015 Paper 2, Q11(b)

WAEC 2015 · Paper 2 · Q11 (b)

In the diagram, $WY$ and $WZ$ are straight lines; $O$ is the centre of circle $WXM$ and $\angle XWM = 48^\circ$ . Calculate the value of $\angle WYZ$ .

∠XWM = 48°measured ∠WYZ = 42°

Mark what you know
$\angle XWM = 48^\circ$ , and O, the centre of the small circle, lies on WM. W, M and Z are on one straight line, and so are W, X and Y.
Think first. Nothing connects the two circles yet. Which two points could you join to link them?
Join XM
Joining X to M makes triangle WXM inside the first circle and the quadrilateral XYZM inside the second. This extra line is the key step.
Think first. W, O and M lie on one line through the centre. What does that make WM, and what does it tell you about $\angle WXM$ ?
Angle in a semicircle
WM goes through the centre O, so it is a diameter.
$\angle WXM = 90^\circ \quad \text{(angle in a semicircle)}$
Think first. You now know two angles of triangle WXM. What is the third?
Angles of triangle WXM
$\angle WMX = 180^\circ - 90^\circ - 48^\circ = 42^\circ$
Think first. X, Y, Z and M are on the second circle. Where is $\angle WMX$ in relation to XYZM?
Exterior angle of a cyclic quadrilateral
XYZM is cyclic. W, M and Z are in line, so $\angle WMX$ is the exterior angle at M. It equals the interior opposite angle, at Y:
$\angle WYZ = \angle XYZ = \angle WMX = 42^\circ$
Answer, and test it
$\angle WYZ = 42^\circ$ .
Now drag Z along the line, or change $\angle XWM$ with the slider. The measured $\angle WYZ$ is always $90^\circ - \angle XWM$ , wherever Z is. The reasons, not the picture, make it true.

WAEC 2012: your turn, one step at a time

This time you find each angle. Answer a step, then press Next step for the next one.

Guided problem · WAEC 2012 Paper 2, Q3(a)

WAEC 2012 · Paper 2 · Q3 (a)

In the diagram, $\overline{TU}$ is a tangent to the circle. $\angle RVU = 100^\circ$ and $\angle URS = 36^\circ$ . Calculate the value of angle $STU$ .

Spot the cyclic quadrilateral
R, V, U and S all lie on the circle, so RVUS is a cyclic quadrilateral. R, S and T are in a straight line, so $\angle UST$ is an exterior angle of RVUS.
Check yourself
Find $\angle UST$ .
Answer in degrees
Use the tangent
TU is a tangent at U, and US is a chord from the point of contact.
Check yourself
Find $\angle SUT$ .
Answer in degrees
Finish in triangle SUT
Check yourself
Find $\angle STU$ .
Answer in degrees

One more theory question

This one mixes algebra with two circle theorems. Write each theorem as an equation, then solve the pair of equations together.

WAEC 2022 · Paper 2 · Q4

Circle geometry

In the diagram, $BCDE$ is a circle with centre $A$ . $\angle BCD = (2x + 40)^\circ$ , $\angle BAD = (5x - 35)^\circ$ , $\angle BED = (2y + 10)^\circ$ and $\angle ADC = 40^\circ$ . Find:

(a)
the values of $x$ and $y$ ;
Separate values with commas, e.g. 3, −2
(b)
$\angle ABC$ .

Worked solution (try it first)

(a)

$BCDE$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$ : $(2x + 40) + (2y + 10) = 180$ , which gives $x + y = 65$ .
The angle at the centre is twice the angle at the circumference on the same arc $BD$ : $5x - 35 = 2(2y + 10)$ , which gives $5x - 4y = 55$ .
From the first equation, $x = 65 - y$ .
Substitute: $5(65 - y) - 4y = 55$ , so $325 - 9y = 55$ , $9y = 270$ and $y = 30$ .
Then $x = 35$ .

(b)

With $x = 35$ : $\angle BCD = 110^\circ$ and $\angle BAD = 140^\circ$ .
The angles of quadrilateral $ABCD$ add up to $360^\circ$ .
So $\angle ABC = 360^\circ - 140^\circ - 110^\circ - 40^\circ$
$= 70^\circ$ .

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