Angle at the Center & The Semicircle Theorem
Every arc of a circle subtends angles both at the center of the circle and at any point along the remaining circumference.
The angle subtended by an arc at the center of a circle is twice the angle subtended by it at any point on the circumference: $$\angle AOB = 2 \times \angle APB$$
Proof Sketch (Deductive Reasoning): Draw the radius $OP$. Triangle $OAP$ is isosceles ($OA = OP = r$), so $\angle OAP = \angle OPA = x$. Triangle $OBP$ is also isosceles ($OB = OP = r$), so $\angle OBP = \angle OPB = y$. The exterior angle of $\triangle OAP$ at $O$ is $2x$, and for $\triangle OBP$ it is $2y$. Hence: $$\angle AOB = 2x + 2y = 2(x + y) = 2\angle APB$$
If $AB$ is a diameter passing through center $O$, the angle at the center is a straight angle ($180^\circ$). Therefore, the angle subtended at the circumference is always a right angle: $$\angle APB = 90^\circ$$
The angle at the center ($2\theta$) is double the angle on the circumference ($\theta$) subtended by arc $AB$.
Angles in the Same Segment & Cyclic Quadrilaterals
When multiple angles are subtended by the same chord or arc into the same segment, they share equal angular measures.
Angles subtended by the same arc (or chord) at the circumference in the same segment are equal: $$\angle APB = \angle AQB$$
A quadrilateral whose four vertices all lie on a circle is called a cyclic quadrilateral.
- Opposite angles are supplementary: $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$.
- Exterior angle property: An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
In any cyclic quadrilateral, opposite pairs of vertices sum to $180^\circ$.
Tangents, Radii & The Alternate Segment Theorem
A tangent is a straight line that touches the circle at exactly one point (point of tangency).
- Perpendicular Radius: The tangent at any point of a circle is perpendicular to the radius through the point of contact ($OT \perp PT \implies \angle OTP = 90^\circ$).
- Tangents from an External Point: Tangents drawn to a circle from an external point $P$ to points of contact $T_1$ and $T_2$ are equal in length: $PT_1 = PT_2$.
- The line connecting $P$ to center $O$ bisects $\angle T_1 P T_2$ and $\angle T_1 O T_2$.
The angle between a tangent and a chord drawn through the point of contact is equal to the angle subtended by the chord in the alternate segment: $$\angle TAB = \angle ACB$$
Alternate Segment Theorem: The angle between tangent and chord $AB$ equals the inscribed angle $\angle ACB$ in the alternate segment.
Chords & Intersecting Chords Theorem (Power of a Point)
Chords possess important symmetry properties and proportional relationships when they intersect inside or outside the circle.
A line drawn from the center of a circle perpendicular to a chord bisects the chord: $$OM \perp AB \implies AM = MB = \frac{1}{2}AB$$ Conversely, the perpendicular bisector of any chord passes through the center $O$.
- Internal Intersection: If chords $AB$ and $CD$ intersect at point $P$ inside the circle: $$PA \times PB = PC \times PD$$
- External Secant Intersection: If secants $PAB$ and $PCD$ meet at external point $P$: $$PA \times PB = PC \times PD$$
- Tangent-Secant Theorem: If a secant $PAB$ and tangent $PT$ meet at external point $P$: $$PT^2 = PA \times PB$$
Worked Example 1: Multi-Theorem Angle Deduction
In a circle with center $O$, points $A$, $B$, and $C$ lie on the circumference. Line $TA$ is a tangent to the circle at point $A$. Given that $\angle TAB = 52^\circ$ and $\angle OBC = 28^\circ$, calculate:
a) $\angle ACB$
b) $\angle AOB$
c) $\angle BAC$
Step-by-step Solution:
Part a: By the Alternate Segment Theorem, the angle between the tangent $TA$ and the chord $AB$ equals the angle subtended by $AB$ in the alternate segment:
$\angle ACB = \angle TAB = 52^\circ$ [Criterion A: Knowledge & Understanding]
Part b: The angle subtended by arc $AB$ at the center $O$ is double the angle subtended at the circumference by the same arc:
$\angle AOB = 2 \times \angle ACB = 2 \times 52^\circ = 104^\circ$
Part c: In $\triangle OBC$, $OB = OC$ (both are radii of the circle), so $\triangle OBC$ is isosceles.
Therefore, $\angle OCB = \angle OBC = 28^\circ$.
Now, $\angle ACO = \angle ACB - \angle OCB = 52^\circ - 28^\circ = 24^\circ$.
In $\triangle OAC$, $OA = OC$ (both radii), so $\triangle OAC$ is isosceles $\implies \angle OAC = \angle ACO = 24^\circ$.
In $\triangle OAB$, $OA = OB$ (both radii), so $\triangle OAB$ is isosceles with vertex angle $\angle AOB = 104^\circ$.
Base angles: $\angle OAB = \frac{180^\circ - 104^\circ}{2} = \frac{76^\circ}{2} = 38^\circ$.
Therefore, $\angle BAC = \angle OAB + \angle OAC = 38^\circ + 24^\circ = 62^\circ$. [Criterion C: Clear Mathematical Communication]
Worked Example 2: Extended Tangent-Secant Theorem
A tangent $PT$ of length $12\text{ cm}$ touches a circle at $T$. A secant line from $P$ passes through the circle, intersecting it at points $A$ and $B$, where $A$ lies between $P$ and $B$. If $PA = 8\text{ cm}$, determine the length of chord $AB$.
Step-by-step Solution:
By the Tangent-Secant Theorem: $$PT^2 = PA \times PB$$ Substitute known lengths: $$12^2 = 8 \times PB$$ $$144 = 8 \times PB \implies PB = \frac{144}{8} = 18\text{ cm}$$ Since $P$, $A$, and $B$ are collinear and $A$ lies between $P$ and $B$: $$PB = PA + AB \implies 18 = 8 + AB \implies AB = 10\text{ cm}$$ [Criterion A: Accurate Computation]
Tiered Practice Exercises
10 carefully sequenced problems advancing from foundational angle theorem applications to complex proofs and algebraic geometric modelling.
Points $A$, $B$, and $C$ lie on a circle with center $O$. If $\angle AOB = 136^\circ$, find the size of $\angle ACB$ subtended at the circumference.
View Solution & Markscheme
Method & Marks:
The angle subtended at the circumference is half the angle subtended at the center: [M1]
$\angle ACB = \frac{1}{2} \times \angle AOB = \frac{1}{2} \times 136^\circ = 68^\circ$. [A1]
In a circle, $AB$ is a diameter and $C$ is a point on the circumference. If $\angle ABC = 38^\circ$, calculate $\angle BAC$.
View Solution & Markscheme
Method & Marks:
By the angle in a semicircle theorem, $\angle ACB = 90^\circ$. [M1]
The sum of angles in $\triangle ABC$ is $180^\circ$:
$\angle BAC = 180^\circ - 90^\circ - 38^\circ = 52^\circ$. [A1]
Quadrilateral $PQRS$ is cyclic. If $\angle P = (2x + 15)^\circ$ and $\angle R = (3x - 10)^\circ$, find the value of $x$ and the measure of angle $P$.
View Solution & Markscheme
Method & Marks:
Opposite angles in a cyclic quadrilateral sum to $180^\circ$: [M1]
$(2x + 15) + (3x - 10) = 180$
$5x + 5 = 180 \implies 5x = 175 \implies x = 35$. [A1]
$\angle P = 2(35) + 15 = 70 + 15 = 85^\circ$. [A1]
Tangents $PA$ and $PB$ are drawn to a circle with center $O$ from an external point $P$. If $\angle APB = 46^\circ$, find the measure of the reflex angle $\angle AOB$.
View Solution & Markscheme
Method & Marks:
Tangents are perpendicular to radii: $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$. [M1]
In quadrilateral $OAPB$, angles sum to $360^\circ$:
$\angle AOB = 360^\circ - 90^\circ - 90^\circ - 46^\circ = 180^\circ - 46^\circ = 134^\circ$. [M1]
Reflex $\angle AOB = 360^\circ - 134^\circ = 226^\circ$. [A1]
A line $XY$ is tangent to a circle at $B$. Chord $AB$ creates an angle $\angle ABX = 64^\circ$. Point $C$ is on the major arc $AB$. Find $\angle ACB$ and state the theorem used.
View Solution & Markscheme
Method & Marks:
By the Alternate Segment Theorem, the angle between the tangent $XY$ and chord $AB$ equals the angle subtended by chord $AB$ in the alternate segment: [M1][R1]
$\angle ACB = \angle ABX = 64^\circ$. [A1]
A circle has center $O$ and radius $13\text{ cm}$. A chord $CD$ is located $5\text{ cm}$ away from the center (perpendicular distance). Calculate the length of the chord $CD$.
View Solution & Markscheme
Method & Marks:
Let $M$ be the midpoint of $CD$. $OM \perp CD$ and $OM = 5\text{ cm}$, $OC = 13\text{ cm}$. [M1]
In right-angled triangle $\triangle OMC$, by Pythagoras:
$MC^2 = OC^2 - OM^2 = 13^2 - 5^2 = 169 - 25 = 144 \implies MC = 12\text{ cm}$. [A1]
Since the perpendicular from the center bisects the chord, $CD = 2 \times MC = 2 \times 12 = 24\text{ cm}$. [A1]
Two chords $AB$ and $CD$ intersect inside a circle at point $X$. If $AX = 4\text{ cm}$, $XB = 9\text{ cm}$, and $CX = 3\text{ cm}$, find the length of segment $XD$ and the total length of chord $CD$.
View Solution & Markscheme
Method & Marks:
By the Intersecting Chords Theorem: [M1]
$AX \times XB = CX \times XD$
$4 \times 9 = 3 \times XD \implies 36 = 3 \times XD \implies XD = 12\text{ cm}$. [A1]
Total chord length $CD = CX + XD = 3 + 12 = 15\text{ cm}$. [A1]
A secant line from an external point $P$ intersects a circle at $A$ and $B$ (with $PA < PB$). A second secant from $P$ passes through the center $O$ of the circle, intersecting it at $C$ and $D$. If $PA = 4\text{ cm}$, $AB = 8\text{ cm}$, and the circle has radius $r = 5\text{ cm}$, determine the distance $PO$ from point $P$ to the center of the circle.
View Solution & Markscheme
Method & Marks:
Let $d = PO$. The secant through the center intersects the circle at $C$ (near point) and $D$ (far point):
$PC = d - r = d - 5$ and $PD = d + r = d + 5$. [M1]
Total length $PB = PA + AB = 4 + 8 = 12\text{ cm}$.
By the Intersecting Secants Theorem (Power of a Point): [M1]
$PA \times PB = PC \times PD$
$4 \times 12 = (d - 5)(d + 5)$
$48 = d^2 - 25 \implies d^2 = 73 \implies d = \sqrt{73} \approx 8.54\text{ cm}$. [A1][R1]
Points $A, B, C, D$ lie consecutively on a circle. Chords $AC$ and $BD$ intersect at right angles at $M$. Prove that $MA^2 + MB^2 + MC^2 + MD^2 = 4R^2$, where $R$ is the radius of the circle.
View Solution & Markscheme
Method & Marks:
Let diameter $AA'$ be drawn through $A$. Then $\angle A'CA = 90^\circ$ (angle in semicircle), so $AC \perp A'C$. [M1]
Since $BD \perp AC$, $A'C$ is parallel to $BD$.
In a circle, parallel chords trap congruent arcs, so chord $A'D = BC$. [M1]
In right-angled triangle $\triangle ACA'$, by Pythagoras: $AC^2 + A'C^2 = (2R)^2 = 4R^2$. [M1]
Substitute $A'C^2$ via right-angled triangle $\triangle A'CD$: $A'D^2 = BC^2$.
Also, in right triangles at $M$ with perpendicular chords:
$AB^2 = MA^2 + MB^2$, $BC^2 = MB^2 + MC^2$, $CD^2 = MC^2 + MD^2$, $DA^2 = MD^2 + MA^2$.
Summing these pairs reveals: $MA^2 + MB^2 + MC^2 + MD^2 = 4R^2$. [A1][R1]
Criterion B & D Investigation: An architect is designing a circular observation rotunda. Two sightlines from an observation tower $P$ outside the perimeter intersect the rotunda: one is tangent to the eastern edge at $T$, and the other passes through the center doors at $A$ and exits at $B$. If $PA = x$, $AB = 2x + 1$, and $PT = x + 4$, find $x$ and determine the rotunda's internal diameter $AB$.
View Solution & Markscheme
Method & Marks:
Total secant length $PB = PA + AB = x + (2x + 1) = 3x + 1$. [M1]
By the Tangent-Secant Theorem: $PT^2 = PA \times PB$ [M1]
$(x + 4)^2 = x(3x + 1)$
$x^2 + 8x + 16 = 3x^2 + x$
Rearrange into a quadratic equation: $2x^2 - 7x - 16 = 0$. [M1]
Using the quadratic formula: $x = \frac{7 \pm \sqrt{(-7)^2 - 4(2)(-16)}}{2(2)} = \frac{7 \pm \sqrt{49 + 128}}{4} = \frac{7 \pm \sqrt{177}}{4}$.
Since distance must be positive: $x = \frac{7 + 13.304}{4} = \frac{20.304}{4} \approx 5.076\text{ m}$. [A1]
The diameter $AB = 2x + 1 = 2(5.076) + 1 \approx 11.15\text{ m}$. [A1][R1: Contextual reflection]
Complete Circle Theorems Reference Sheet
Angle at Center
$$\angle AOB = 2 \times \angle APB$$
Subtended by the same arc.Angle in Semicircle
$$\angle APB = 90^\circ$$
When $AB$ is the diameter.Angles in Same Segment
$$\angle APB = \angle AQB$$
Subtended by the same chord.Cyclic Quadrilateral
$$\angle A + \angle C = 180^\circ$$
Opposite angles are supplementary.Tangent & Radius
$$OT \perp PT \iff \angle OTP = 90^\circ$$
Perpendicular at contact point.Alternate Segment
$$\angle TAB = \angle ACB$$
Angle between tangent & chord.Intersecting Chords
$$PA \cdot PB = PC \cdot PD$$
Internal intersection point $P$.Tangent-Secant Theorem
$$PT^2 = PA \cdot PB$$
External point $P$, tangent $PT$.Top 5 Common MYP Circle Pitfalls
- Confusing Central Angle with Circumference Angle: Students often set $\angle AOB = \frac{1}{2}\angle APB$ instead of $\angle AOB = 2\angle APB$. Remember: the central angle is always wider and opens more broadly than the angle at the circumference.
- Assuming Any 4-Sided Shape Inside a Circle is Cyclic: All four vertices must lie directly on the circumference. If one vertex is at the center $O$ (e.g., $OABC$), it is NOT a cyclic quadrilateral and opposite angles do NOT sum to $180^\circ$.
- Wrong Segments in Intersecting Chords / Secants: When applying $PA \cdot PB = PC \cdot PD$ externally, students often use $AB$ instead of the total secant length $PB = PA + AB$. Always measure all segments from the point of intersection $P$ to the circle!
- Misidentifying the Alternate Segment: The angle between the tangent and chord $AB$ equals the angle inside the triangle subtended by $AB$ on the opposite (alternate) side, not the adjacent angle in the same segment.
- Missing Hidden Isosceles Triangles: Any triangle formed by two radii from center $O$ to circumference points (e.g., $\triangle OAB$) is automatically an isosceles triangle ($OA = OB = r$). Always mark these equal sides immediately.
Self-Grading Mini Test
5 quiz questions on circle theorems. Choose an answer to see feedback and a worked solution.
In a circle with center $O$, points $A$ and $B$ lie on the circumference. If $\angle APB = 44^\circ$, what is the size of the central angle $\angle AOB$?
Explanation: The central angle is twice the inscribed angle on the circumference subtended by the same arc: $\angle AOB = 2 \times 44^\circ = 88^\circ$.
In a cyclic quadrilateral $ABCD$, $\angle A = 73^\circ$ and $\angle B = 105^\circ$. What is the measure of angle $C$?
Explanation: Opposite angles of a cyclic quadrilateral sum to $180^\circ$. Therefore, $\angle C = 180^\circ - \angle A = 180^\circ - 73^\circ = 107^\circ$.
Segment $AB$ is the diameter of a circle. Point $C$ lies on the circumference. If $\angle CAB = 32^\circ$, what is the measure of $\angle ABC$?
Explanation: Because $AB$ is the diameter, $\angle ACB = 90^\circ$. In right triangle $\triangle ABC$, $\angle ABC = 180^\circ - 90^\circ - 32^\circ = 58^\circ$.
Chords $AB$ and $CD$ intersect inside a circle at $P$. If $PA = 3\text{ cm}$, $PB = 8\text{ cm}$, and $PC = 4\text{ cm}$, what is the length of $PD$?
Explanation: By the intersecting chords theorem: $PA \times PB = PC \times PD \implies 3 \times 8 = 4 \times PD \implies 24 = 4 \times PD \implies PD = 6\text{ cm}$.
From an external point $P$, tangent $PT = 6\text{ cm}$ touches a circle at $T$. A secant line from $P$ cuts the circle at $A$ and $B$. If $PA = 4\text{ cm}$, what is the length of chord $AB$?
Explanation: By the Tangent-Secant Theorem: $PT^2 = PA \times PB \implies 6^2 = 4 \times PB \implies 36 = 4 \times PB \implies PB = 9\text{ cm}$. Then $AB = PB - PA = 9 - 4 = 5\text{ cm}$.