Grade 9

Circles

The set of points at fixed distance from a center, with its tangents, chords and inscribed angles.

IntuitionOne center, constant radius, and the surprise of constant angles

Pin one end of a taut string of length $R > 0$ at a fixed point $O$ and sweep the other end around the plane: the curve traced out is the circle $(O, R)$ , the most symmetric figure in plane geometry. Every segment joining two points on the circle is a chord $AB$ , and a line that touches the circle at a single point is a tangent. Hidden inside this simple definition is a remarkable geometric surprise: if you stand at any point $C$ on the major arc and look at the chord, your viewing angle (the inscribed angle) is always exact half of the central angle $\angle AOB$ , namely $\angle ACB = \dfrac{1}{2}\angle AOB$ — no matter where you move along that arc! In particular, when the chord is a diameter ( $180^\circ$ central angle), Thales' theorem guarantees a right angle $\angle ACB = 90^\circ$ everywhere on the semicircle. Use the interactive unit circle below to explore how rotating a point around the center changes its central angle and coordinates.

A point rotating on the unit circle around center

O

at central angle

\theta

, with coordinates

(\cos\theta, \sin\theta)

SchoolChords, tangents, inscribed angles, and power of a point

Definition: Circle, chords, tangents, and inscribed angles

The circle $(O, R)$ is the set of all points in the plane whose distance from the center satisfies $OM = R$ . A line through the center perpendicular to a chord bisects that chord ( $OM \perp AB \iff MA = MB$ ), and a tangent line at the contact point is perpendicular to the radius ( $OT \perp d$ ). Any inscribed angle $\angle ACB$ equals half the central angle $\angle AOB$ subtending the same arc:

\angle ACB = \frac{1}{2}\angle AOB

When the chord is a diameter, the central angle is $\angle AOB = 180^\circ$ , so Thales' semicircle theorem gives $\angle ACB = 90^\circ$ for every point on the semicircle. Furthermore, whenever two lines through a point $P$ (either inside or outside the circle) meet the circle along chords $AB$ and $CD$ , the product of the segment lengths from the intersection point is constant (power of a point):

PA \cdot PB = PC \cdot PD

Fundamental theorems of circle geometry
Configuration	Geometric condition	Key relation
Perpendicular from center to chord $AB$	Foot $M$ on chord	$OM \perp AB \iff MA = MB$ , $R^2 = OM^2 + MA^2$
Tangents from external point $P$	Contact points $T_1, T_2$ on circle	$OT_1 \perp PT_1$ and $PT_1 = PT_2$
Inscribed angle on arc $AB$	Vertex $C$ on circle, center $O$	$\angle ACB = \dfrac{1}{2}\angle AOB$
Thales' semicircle on diameter $AB$	$AB = 2R$ passes through $O$	$\angle ACB = 90^\circ$
Power of a point $P$	Lines $PAB$ and $PCD$ meet circle	$PA \cdot PB = PC \cdot PD$

UndergraduateTwo key theorems and their proofs

Theorem: Inscribed Angle Theorem and Thales' Semicircle Corollary

For any three points on a circle $(O, R)$ , the inscribed angle equals half the central angle subtending the same arc: $\angle ACB = \dfrac{1}{2}\angle AOB$ . In particular, when the chord $AB$ is a diameter, every inscribed angle subtending it is a right angle ( $\angle ACB = 90^\circ$ ).

Why is it true?

Because all radii of a circle have the exact same length, joining the center to the three vertices splits the configuration into isosceles triangles; the exterior angle theorem on those isosceles triangles immediately doubles each half of the inscribed angle into the corresponding part of the central angle.

Proof

Step 1 (draw the diameter from the vertex and use isosceles triangles). Draw the diameter $CD$ passing through the vertex and the center. Because two radii have equal length $OA = OC = R$ , the triangle $\triangle OAC$ is isosceles at the center, so its base angles are equal: $\angle OCA = \angle OAC$ .

Step 2 (apply the exterior angle theorem to both halves). The exterior angle at the center equals the sum of the two opposite interior angles, giving $\angle AOD = \angle OCA + \angle OAC = 2\angle OCA$ . Applying the exact same isosceles-triangle argument to $\triangle OBC$ gives $\angle BOD = 2\angle OCB$ .

Step 3 (combine the two halves and deduce Thales' theorem). Adding the two central angles (or subtracting them when the center lies outside the inscribed angle) yields $\angle AOB = \angle AOD + \angle BOD = 2(\angle OCA + \angle OCB) = 2\angle ACB$ . In the special case where the chord is a diameter, the central angle is straight ( $\angle AOB = 180^\circ$ ), which immediately gives Thales' semicircle theorem $\angle ACB = \dfrac{1}{2} \cdot 180^\circ = 90^\circ$ .

Theorem: Power of a Point (Intersecting Chords and Secants)

If two lines through a point $P$ intersect a circle at the endpoints of chords $AB$ and $CD$ , then the products of the segment lengths from that point are equal: $PA \cdot PB = PC \cdot PD$ . When the point lies outside the circle and a tangent segment to the circle is drawn, its squared length satisfies $PT^2 = PA \cdot PB$ .

Why is it true?

Although a circle is curved, the inscribed angle theorem locks the angles of the two triangles formed by the intersecting chords into equality, making the triangles similar; converting their side-length proportion into a cross-product turns a ratio of lengths into an invariant product.

Proof

Step 1 (form two triangles using the chord endpoints). Join the endpoints by segments $AC$ and $BD$ to form the two triangles $\triangle PAC$ and $\triangle PDB$ .

Step 2 (prove the two triangles are similar via inscribed angles). Because inscribed angles subtending the same arc are equal, we have $\angle CAP = \angle BDP$ . In addition, the angles at the intersection point are equal (as vertical angles when the point is inside the circle, or as the same shared angle when the point is outside): $\angle APC = \angle DPB$ . By the Angle-Angle similarity criterion, $\triangle PAC \sim \triangle PDB$ .

Step 3 (take the ratio of corresponding sides and cross-multiply). Similarity of the two triangles gives the side ratio $\dfrac{PA}{PD} = \dfrac{PC}{PB}$ . Cross-multiplying both sides yields $PA \cdot PB = PC \cdot PD$ .

UndergraduateReal-World Applications and Worked Examples

Circle geometry underpins civil engineering, navigation, optics, and astronomy. Archaeologists and machinists reconstruct the radius of a broken circular wheel, pipe, or ceramic plate from a single shard by measuring a chord and its perpendicular bisector $OM \perp AB$ together with the Pythagorean relation $R^2 = OM^2 + MA^2$ . Before GPS, coastal navigators used the inscribed angle theorem $\angle ACB = \dfrac{1}{2}\angle AOB$ to steer a ship along a safe circle of constant angle between two lighthouses (and used Thales' semicircle $\angle ACB = 90^\circ$ to check whether the ship had crossed the danger circle around a submerged reef). On a spherical Earth or in satellite communications, the tangent-secant power formula $PT^2 = PA \cdot PB$ gives the exact distance to the visible horizon from a coastal tower or spacecraft at a given altitude.

Example: Finding the width of a water surface in a circular pipe

In a circular water pipe of cross-section $(O, R)$ with radius $R = 10$ cm, the flat water surface forms a horizontal chord $AB$ whose perpendicular distance from the center is $OM = 6$ cm. Find the exact width of the water surface.

Solution

Step 1: bisect the chord using the perpendicular from the center. Because the radius segment from the center meets the chord at a right angle ( $OM \perp AB$ ), its foot is the midpoint of the chord, so $MA = MB$ .

Step 2: apply the Pythagorean theorem in right triangle $\triangle OMA$ . Since the hypotenuse is the radius and one leg is the distance to the center, the half-chord satisfies $MA^2 = R^2 - OM^2 = 10^2 - 6^2 = 100 - 36 = 64$ .

Step 3: double the half-chord to find the full width. Taking the square root gives $MA = \sqrt{64} = 8$ cm, so the full width of the water surface is $AB = 2 \cdot MA = 2 \cdot 8 = 16$ cm.

Example: Using intersecting chords to find a missing walkway segment

Inside a circular plaza, two straight brick walkways form chords $AB$ and $CD$ that cross at point $P$ . A surveyor measures $PA = 4$ m, $PB = 9$ m, and $PC = 6$ m. Find the length of the remaining walkway segment $PD$ .

Solution

Step 1: write the intersecting chords equation. By the power of a point theorem for two chords crossing inside a circle, their segment lengths satisfy $PA \cdot PB = PC \cdot PD$ .

Step 2: substitute the three known lengths. Plugging the given values into the equation gives $4 \cdot 9 = 6 \cdot PD$ , which simplifies to $36 = 6 \cdot PD$ .

Step 3: solve for the unknown segment. Dividing both sides by the coefficient gives $PD = \dfrac{36}{6} = 6$ m.

Historical note

Ancient Greek tradition credits Thales of Miletus (6th century BCE) with proving that any angle inscribed in a semicircle is a right angle ( $\angle ACB = 90^\circ$ ). Around 300 BCE, Euclid compiled the complete classical theory of circles in Book III of the Elements, including the general inscribed angle theorem $\angle ACB = \dfrac{1}{2}\angle AOB$ (Proposition 20) and the intersecting chords and tangent-secant theorems $PA \cdot PB = PC \cdot PD$ (Propositions 35–36). Shortly afterward, Archimedes bounded the circumference-to-diameter ratio of a circle by inscribing and circumscribing regular polygons up to 96 sides, proving $3\dfrac{10}{71} < \pi < 3\dfrac{1}{7}$ .

Euclid of Alexandria Archimedes of Syracuse

Points $A$ , $B$ , $C$ lie on a circle with center $O$ . If the central angle subtending arc $AB$ is $\angle AOB = 110^\circ$ , what is the measure of the inscribed angle $\angle ACB$ subtending the same arc?

In a circle of radius $R = 13$ cm, a chord has length $AB = 24$ cm. What is the perpendicular distance $OM$ from the center $O$ to the chord?

Segment $AB$ is a diameter of a circle, and $C$ is a point on the circle such that $\angle CAB = 35^\circ$ . What is the measure of $\angle ABC$ ?

From an external point $P$ , a tangent $PT$ of length $PT = 6$ cm and a secant $PAB$ are drawn to a circle, with $PA = 4$ cm. What is the distance $PB$ from $P$ to the far intersection point?

References

Euclid (trans. Thomas L. Heath) (1956). Euclid's Elements (Books I–XIII)
H. S. M. Coxeter, S. L. Greitzer (1967). Geometry Revisited