Question: The circle $ABC$ has centre $O$. $QA$ and $QB$ are tangents from the external point $Q$, and $\angle BAQ = 63°$. Find $\angle AQB$, $\angle AOB$ and $\angle ACB$.

[diagram in original sheet: circle with tangents QA, QB and C on the major arc]

Step 1. Tangents from an external point are equal: $AQ = BQ$, so $\triangle AQB$ is isosceles with base angles $63°$:

$$\angle AQB = 180° - 63° - 63° = 54°$$

Step 2. A tangent meets its radius at $90°$: $\angle OAQ = \angle OBQ = 90°$ (tan $\perp$ rad).

Angles of quadrilateral $OAQB$ sum to $360°$:

$$\angle AOB = 360° - 90° - 90° - 54° = 126°$$

Step 3. The angle at the centre is twice the angle at the circumference, standing on the same arc $AB$:

$$\angle ACB = \dfrac{126°}{2} = 63°$$

Step 4. Quote a property for every line of working — "tangents from an external point", "tan $\perp$ rad", "$\angle$ at centre $= 2\angle$ at circumference" — the reasons carry marks.

⚠ Watch out: The right angle sits at the point of contact: $\angle OAQ = 90°$, not $\angle AQO$. And check which arc the circumference angle stands on — a $C$ on the minor arc would see half the reflex $\angle AOB$ instead.

Question: A circle passes through $A$, $B$, $C$ and $D$. The chords $AC$ and $BD$ cross at $P$. Given $\angle ADB = 32°$, $\angle ACD = 46°$ and $\angle CBD = 64°$, find $\angle ACB$, $\angle CPD$ and $\angle BAD$.

[diagram in original sheet: cyclic quadrilateral ABCD with diagonals meeting at P]

Step 1. $\angle ACB$ and $\angle ADB$ stand on the same chord $AB$, on the same side:

$$\angle ACB = \angle ADB = 32° \quad (\angle\text{s in the same segment})$$

Step 2. For $\angle CPD$, work inside $\triangle BPC$ first:

$$\angle BPC = 180° - 64° - 32° = 84° \quad (\angle \text{ sum of } \triangle)$$ $$\angle CPD = 180° - 84° = 96° \quad (\angle\text{s on a straight line})$$

Step 3. $ABCD$ is a cyclic quadrilateral, so opposite angles add to $180°$. The whole angle at $C$ is

$$\angle BCD = \angle ACB + \angle ACD = 32° + 46° = 78°$$ $$\angle BAD = 180° - 78° = 102° \quad (\angle\text{s in opposite segments})$$

⚠ Watch out: Same-segment pairs must stand on the same chord and the same side of it — $\angle ACB$ pairs with $\angle ADB$ (both on $AB$), never with $\angle ADC$. And "opposite segments" means opposite corners of the cyclic quadrilateral: $\angle BAD + \angle BCD$, not two adjacent angles.

Question: $AB$ is a diameter of circle $ACBD$, and $AE$ is the tangent at $A$. The line through $E$, $D$, $C$ cuts the circle at $D$ and $C$, with $\angle BAC = 37°$ and $\angle DCA = 29°$. Find $\angle ABD$, $\angle ADC$, $\angle DAE$ and $\angle AEC$.

[diagram in original sheet: circle with diameter AB, tangent AE at A, and line EDC cutting the circle]

Step 1. $\angle ABD$ and $\angle ACD$ stand on the same chord $AD$:

$$\angle ABD = \angle ACD = 29° \quad (\angle\text{s in same segment})$$

Step 2. $AB$ is a diameter, so $\angle ADB = 90°$ (rt $\angle$ in semicircle). Also $\angle BDC = \angle BAC = 37°$ (same segment, chord $BC$):

$$\angle ADC = 90° - 37° = 53°$$

Step 3. $\angle BAE = 90°$ (tan $\perp$ rad). In $\triangle ABD$: $\angle BAD = 180° - 90° - 29° = 61°$, so

$$\angle DAE = 90° - 61° = 29°$$

Step 4. In $\triangle AEC$: $\angle ACE = 29°$ and $\angle EAC = \angle EAB + \angle BAC = 90° + 37° = 127°$:

$$\angle AEC = 180° - 29° - 127° = 24°$$

⚠ Watch out: The $90°$ from a semicircle needs an actual diameter — check the line passes through the centre before quoting it. Two different right angles live here ($\angle ADB$ from the semicircle, $\angle BAE$ from the tangent): keep them apart.

Question: $AC$ is the diameter of a circle, centre $O$, through $A$, $B$, $C$, $D$ (in order). $\angle BOA = 76°$ and $\angle ACD = 18°$. Stating reasons, find $\angle ABD$, $\angle DBC$ and $\angle BAD$.

[diagram in original sheet: circle with diameter AC and points B, D on opposite arcs]

Step 1. $\angle ABD$ and $\angle ACD$ stand on the same chord $AD$:

$$\angle ABD = \angle ACD = 18° \quad (\angle\text{s in same segment})$$

Step 2. $AC$ is a diameter, so the whole angle at $B$ is

$$\angle ABC = 90° \quad (\text{rt } \angle \text{ in semicircle})$$ $$\angle DBC = 90° - 18° = 72°$$

Step 3. For $\angle BAD$, use the cyclic quadrilateral $ABCD$: first get $\angle BCA$ from the centre angle on chord $AB$:

$$\angle BCA = \dfrac{76°}{2} = 38° \quad (\angle \text{ at centre} = 2\angle \text{ at circumference})$$ $$\angle BCD = 38° + 18° = 56°$$ $$\angle BAD = 180° - 56° = 124° \quad (\angle\text{s in opposite segments})$$

Step 4. Four theorems, one figure — that is exactly how prelim circle questions are built. Chain them one link at a time, quoting each reason.

⚠ Watch out: Before using the cyclic-quadrilateral rule, assemble the whole corner angle: $\angle BCD = \angle BCA + \angle ACD$. Pairing $\angle BAD$ with only part of the opposite corner ($18°$ or $38°$ alone) silently loses the question.

Question: $AB$ and $CD$ are parallel chords of a circle, centre $O$, radius $16$ cm, on opposite sides of the centre. $AB = 18$ cm and $CD = 12$ cm. Find the shortest distance between the chords.

[diagram in original sheet: circle with the two parallel chords, O between them, perpendicular line XY through O]

Step 1. Drop the perpendicular from $O$ to each chord. The perpendicular from the centre bisects the chord, so it meets $AB$ at its midpoint $X$ and $CD$ at its midpoint $Y$:

$$AX = \dfrac{18}{2} = 9, \qquad CY = \dfrac{12}{2} = 6$$

Step 2. Each half-chord, its perpendicular, and a radius form a right-angled triangle. By Pythagoras:

$$OX = \sqrt{16^2 - 9^2} = \sqrt{175} = 13.2287\ldots$$ $$OY = \sqrt{16^2 - 6^2} = \sqrt{220} = 14.8323\ldots$$

Step 3. The chords are on opposite sides of $O$, so the distances add:

$$XY = 13.2287\ldots + 14.8323\ldots = 28.061\ldots \approx 28.1 \text{ cm} \quad (3 \text{ s.f.})$$

⚠ Watch out: Read the diagram before the last step: chords on opposite sides of the centre give $OX + OY$; chords on the same side give $OY - OX$. And it is the half-chord that goes into Pythagoras — using the full $18$ against radius $16$ gives a negative under the root.

Question: $AB$ and $CD$ are parallel chords of a circle, centre $O$, on opposite sides of the centre. $AB = 48$ cm, $CD = 30$ cm, and the perpendicular distance from $O$ to $AB$ is $OP = 7$ cm. Find the radius of the circle and the distance $PQ$ between the chords.

[diagram in original sheet: circle with parallel chords AB and CD, perpendicular line POQ]

Step 1. The perpendicular from the centre bisects the chord:

$$AP = \dfrac{48}{2} = 24 \text{ cm}$$

Step 2. Radius from the right-angled triangle $OPA$ (Pythagoras):

$$r = OA = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm}$$

Step 3. Now use the same radius on the other chord. $CQ = \dfrac{30}{2} = 15$, so

$$OQ = \sqrt{25^2 - 15^2} = \sqrt{400} = 20 \text{ cm}$$

Step 4. Chords on opposite sides of the centre — add the two distances:

$$PQ = 7 + 20 = 27 \text{ cm}$$

Check. $24^2 + 7^2 = 625 = 25^2$ ✓ and $15^2 + 20^2 = 625$ ✓ — both half-chords sit on the same circle.

⚠ Watch out: Pythagoras runs both ways here: adding the squares when the radius is the unknown (Step 2), subtracting when the radius is known (Step 3). Mixing the two directions is the standard slip in chord questions.