Comprehensive Study Notes on Circle Geometry and Concyclicity

Origins and Natural Occurrences of Geometric Shapes

Humanity has maintained a historical fascination with the shapes present in the natural environment. Early cave art documents geometric patterns including triangles, squares, circles, and ovals. For instance, the cave paintings of Gudahandi in Odisha prominently display numerous geometric designs, with the sun depicted using circular geometry. These mathematical forms were directly inspired by observations of natural phenomena.

Circular shapes manifest across various natural structures. Concentric circles form on water surfaces when raindrops fall. Cross-sections of plant stems and the inflorescence structure of sunflowers exhibit circular geometry. Additionally, both the full moon and the sun during a total solar eclipse present circular outlines.

Mathematical Definition and Fundamental Properties of Circles

When analyzing mathematical shapes such as circles, triangles, and squares, the figures are assumed to be drawn on a two-dimensional plane.

A circle is defined as the set of all points on a two-dimensional plane that are equidistant from a fixed point located on that plane. In set notation and locus terminology, a set of points satisfying a specific geometric condition is termed a locus. Thus, a circle is the locus of points that are equidistant from a given fixed point.

The given fixed point is defined as the centre of the circle. The distance from the centre to any point on the circle is defined as the radius of the circle. For a circle with centre $A$, all points on the plane located at a distance equal to length $AB$ from $A$ constitute a circle with centre $A$ and radius $AB$.

A line segment connecting any two points on a circle is called a chord. For points $B$ and $C$ on a circle with centre $A$, the segment $BC$ is a chord. The angle formed between the line segments joining the endpoints of the chord to the centre is the angle subtended by the chord at the centre, represented as $\angle BAC$.

A chord that passes directly through the centre of a circle is defined as a diameter.

Locating the centre of a circular physical object, such as a paper disc, can be accomplished through paper-folding. By folding a circular sheet of paper in half, a straight line crease is formed representing a diameter. Unfolding and folding the paper along a different line produces a second diameter crease. The unique point of intersection between these two linear creases is the exact centre of the circle.

Rotational Symmetry and Loci

A circle possesses complete rotational symmetry. Rotating a circle about its centre through any arbitrary angle leaves its appearance unchanged. Every line passing through the centre, such as a folded crease along a diameter, represents a line of symmetry. Because infinitely many diameters exist, a circle has an infinite number of lines of symmetry.

By comparison, other regular geometric figures have finite symmetries: a square possesses $4$ lines of symmetry and $4$-fold rotational symmetry; a regular pentagon possesses $5$ lines of symmetry; and a regular hexagon possesses $6$ lines of symmetry.

The diameter represents the longest possible chord in any given circle. In a circle having a radius of $5\,\text{units}$, the length of the longest chord (the diameter) is $10\,\text{units}$. Conversely, as a chord is positioned closer to the boundary edge of the circle away from the centre, its length decreases toward a theoretical lower limit of $0$.

The locus of points equidistant from two fixed given points $A$ and $B$ is the perpendicular bisector of the line segment $AB$. Every point lying on the perpendicular bisector of segment $AB$ is equidistant from both point $A$ and point $B$.

Circle Existence Through Specified Points

For two given points $A$ and $B$ on a plane, infinitely many distinct circles can be drawn passing through both points. The centre of any circle containing both $A$ and $B$ must lie on the perpendicular bisector of segment $AB$.

The midpoint of segment $AB$ serves as the centre for the smallest possible circle passing through points $A$ and $B$. For this smallest circle, segment $AB$ serves as the diameter, and its radius is equal to $\frac{1}{2} AB$. As a point moves further away from segment $AB$ along its perpendicular bisector, that point can serve as the centre of a larger circle passing through $A$ and $B$. Moving further out along the perpendicular bisector increases the required radius without bound, and the arc segment connecting $A$ and $B$ appears progressively less curved (flatter).

Through three points on a plane, circle construction depends entirely on whether the points are collinear:

If three points $A$, $B$, and $C$ are collinear, no circle can pass through all three points. The perpendicular bisectors of segments $AB$ and $BC$ are strictly parallel lines and do not intersect to yield a centre point. Consequently, a straight line can intersect a circle at a maximum of two distinct points.

Theorem 1 states that there is a unique circle passing through three non-collinear points $A$, $B$, and $C$.

To demonstrate Theorem 1: Let a circle pass through non-collinear points $A$, $B$, and $C$ with centre $O$. Because $OA = OB$, the point $O$ must lie on the perpendicular bisector of line segment $AB$. Because $OA = OC$, the point $O$ must also lie on the perpendicular bisector of line segment $AC$. Since points $A$, $B$, and $C$ are non-collinear, line segments $AB$ and $AC$ are non-parallel, meaning their respective perpendicular bisectors are intersecting lines. Two non-parallel lines intersect at exactly one unique point on a plane, fixing $O$ as a unique centre. Drawing a circle centered at $O$ with radius equal to length $OA$ constructs the unique circle passing through $A$, $B$, and $C$.

Three non-collinear points also define a triangle $\Delta ABC$. The unique circle passing through the vertices of a triangle is called the circumcircle of the triangle, and its centre $O$ is called the circumcentre. The circumcentre is constructed by finding the intersection point of the perpendicular bisectors of the triangle's sides. The relative position of the circumcentre depends on the angles of the triangle:

For an acute-angled triangle, the circumcentre $O$ lies inside the triangle.

For an obtuse-angled triangle, the circumcentre $O$ lies outside the triangle.

For a right-angled triangle, the circumcentre $O$ lies exactly at the midpoint of the hypotenuse.

Subtended Angles and Chord Properties

When a physical cord or segment connects two boundary points on a circular frame, it models a chord. Line segments drawn from the endpoints of this chord to the centre form a central angle subtended by the chord.

Theorem 2 states that equal chords of a circle subtend equal angles at the centre of the circle.

Given two chords $AB$ and $DE$ in a circle with centre $C$ such that $AB = DE$:

To show that $\angle ACB = \angle DCE$:

Line segments $CA$, $CB$, $CD$, and $CE$ are all radii of the circle, establishing $CA = CD = r$ and $CB = CE = r$. Given that $AB = DE$, triangle $\Delta CAB$ is congruent to triangle $\Delta CDE$ by the Side-Side-Side (SSS) congruence criterion ($\Delta CAB \cong \Delta CDE$). By corresponding parts of congruent triangles, $\angle ACB = \angle DCE$.

Theorem 3 represents the converse of Theorem 2 and states that if two chords of a circle subtend equal angles at the centre, then the chords are equal in length.

Given two chords $AB$ and $DE$ in a circle with centre $C$ such that $\angle ACB = \angle DCE$:

To show that $AB = ED$:

Line segments $AC$, $BC$, $FC$, and $DC$ are equal to the radius $r$, giving $AC = DC = r$ and $BC = EC = r$. Because $AC = DC$, $BC = EC$, and the included angles $\angle ACB = \angle DCE$ are equal, triangle $\Delta ACB$ is congruent to triangle $\Delta DCE$ by the Side-Angle-Side (SAS) congruence criterion ($\Delta ACB \cong \Delta DCE$). Therefore, AB = ED$.\n\n# Midpoints and Perpendiculars of Chords\n\nTheorem 4 states that the line segment joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.\n\nGiven a chord AB$with midpoint$M$(such that$AM = BM$) in a circle with centre$C:\n\nTo show that CM \perp AB:\n\nConsider triangles \Delta CMA$and$\Delta CMB$.$CA = CB = r$(radii),$AM = BM$(given midpoint condition), and$CM = CM$(shared side). By the SSS congruence criterion,$\Delta CMA \cong \Delta CMB$. Consequently,$\angle CMA = \angle CMB$. Because$A$,$M$, and$B$lie along a straight line segment,$\angle CMA + \angle CMB = 180^\circ$. Therefore,$\angle CMA = \angle CMB = 90^\circ$, proving$CM$is perpendicular to$AB.\n\nThe converse to Theorem 4 states that the perpendicular line segment dropped from the centre of a circle to a chord bisects the chord.\n\nTo prove this converse: In right-angled triangles \Delta CMA$and$\Delta CMB$,$\angle CMA = \angle CMB = 90^\circ$, hypotenuses$CA = CB = r$, and side$CM = CM$. By the Right angle-Hypotenuse-Side (RHS) congruence criterion,$\Delta CMA \cong \Delta CMB$. Thus,$AM = BM, establishing that the perpendicular bisects the chord.\n\nFor an isosceles triangle \Delta ABC$inscribed in a circle with$AB = AC$, the altitude drawn from vertex$A$to base$BC$passes directly through the centre of the circle because it acts as the perpendicular bisector of chord$BC$.

If two parallel chords of lengths $6\,\text{cm}$ and $8\,\text{cm}$ lie on opposite sides of the centre of a circle of radius $5\,\text{cm}$, the distance between their midpoints is calculated using the Pythagorean relation. For the $6\,\text{cm}$ chord, half-length is $3\,\text{cm}$, giving a perpendicular distance from centre $d_1 = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = 4\,\text{cm}$. For the $8\,\text{cm}$ chord, half-length is $4\,\text{cm}$, giving distance $d_2 = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3\,\text{cm}$. The total separation distance between the midpoints is $d_1 + d_2 = 4\,\text{cm} + 3\,\text{cm} = 7\,\text{cm}$.

Perpendicular Distance of Chords from the Centre

Paper-folding demonstrates chord distance properties: folding a circular paper boundary forms a chord crease, and folding the chord endpoints onto one another creates a perpendicular crease passing through the centre, intersecting the chord at its midpoint.

Theorem 6 states that chords of equal length in a circle are equidistant from the centre of the circle.

Given chords $AB$ and $FG$ in a circle with centre $C$ such that $AB = FG$, with points $E$ and $H$ acting as midpoints of $AB$ and $FG$ respectively, so that $CE \perp AB$ and $CH \perp FG$:

To show that $CE = CH$:

Proof Method 1 (Triangle Congruence): $CA = CF = r$ and $CB = CG = r$. Given $AB = FG$, triangle $\Delta CAB \cong \Delta CFG$ by SSS congruence. Altitudes dropped from corresponding vertices of congruent triangles are congruent; hence CE = CH$.\n\nProof Method 2 (RHS Congruence): Perpendiculars from the centre bisect chords, making AE = \frac{1}{2} AB$and$FH = \frac{1}{2} FG$. Because$AB = FG$,$AE = FH$. In right triangles$\Delta CEA$and$\Delta CHF$,$\angle CEA = \angle CHF = 90^\circ$, hypotenuses$CA = CF = r$, and legs$AE = FH$. By RHS congruence,$\Delta CEA \cong \Delta CHE$, yielding$CE = CH$.

Proof Method 3 (Pythagorean Theorem): Applying the Baudhayana-Pythagoras theorem in right triangle $\Delta CEA$ yields $CA^2 = CE^2 + AE^2$, so $CE = \sqrt{r^2 - AE^2}$. Similarly, $CH = \sqrt{r^2 - FH^2}$. Because $AE = \frac{1}{2} AB = \frac{1}{2} FG = FH$, it follows directly that CE = CH$.\n\nTheorem 7 (Converse of Theorem 6) states that chords of a circle that are equidistant from the centre have equal length.\n\nGiven CE \perp AB$,$CH \perp FG$, and$CE = CH:\n\nTo show that AB = FG:\n\nIn right triangles \Delta CEA$and$\Delta CHF$, hypotenuses$CA = CF = r$and legs$CE = CH$. By RHS congruence,$\Delta CEA \cong \Delta CHE$, so$AE = FH$. Because perpendiculars from the centre bisect the chords,$AB = 2 AE$and$FG = 2 FH$, establishing$AB = FG$.

Theorem 8 addresses unequal chords and states that if $AB$ and $DE$ are two chords of a circle with centre $C$ such that $AB > DE$, then the perpendicular distance from $C$ to $AB$ is less than the perpendicular distance from $C$ to $DE$.

Given chords $AB > DE$ with perpendiculars $CF \perp AB$ and $CG \perp DE$:

To show that $CF < CG$:

In right triangles $\Delta CFA$ and $\Delta CGD$, $\text{CD}^2 = \text{CG}^2 + \text{GD}^2$ and $\text{AC}^2 = \text{CF}^2 + \text{AF}^2$. Because $AC = CD = r$, $\text{CF}^2 + \text{AF}^2 = \text{CG}^2 + \text{GD}^2$. Perpendiculars bisect the chords, so $AF = \frac{1}{2} AB$ and $GD = \frac{1}{2} DE$. Given $AB > DE$, $AF > GD$, which implies $\text{AF}^2 > \text{GD}^2$. Subtracting $\text{AF}^2$ from the equality forces $\text{CF}^2 < \text{CG}^2$, establishing CF < CG$.\n\nThe chord nearest to the centre is the diameter, having a distance of zero and a maximum length of 2r$. As a chord is displaced away from the centre toward the edge, its length decreases. When perpendicular distance$d$equals radius$r$, chord length becomes zero. Generalizing this relation, the length of a chord at perpendicular distance$d$in a circle of radius$r$is given by$2 \sqrt{r^2 - d^2}$. For a circle with radius$r = 7\,\text{cm}$and perpendicular chord distance$d = 6\,\text{cm}$, chord length is$2 \sqrt{7^2 - 6^2} = 2 \sqrt{49 - 36} = 2 \sqrt{13}\,\text{cm}.\n\n# Arc Geometry and Inscribed Angles\n\nAn arc is a connected continuous portion of a circle bounded by two endpoints. For two distinct points A$and$B on a circle:\n\nThe arc covering the shorter boundary path is the minor arc (e.g., arc AXB).\n\nThe arc covering the longer boundary path is the major arc (e.g., arc AYB).\n\nThe angle subtended by an arc at the centre is measured by sweeping along the arc from radius OA$to radius$OB$. A minor arc subtends a central angle$\angle AOB < 180^\circ$. A major arc subtends a reflex central angle$\text{reflex } \angle AOB > 180^\circ$.

For any point $C$ lying on the circle outside a given arc $AXB$, the angle $\angle ACB$ is the angle subtended by arc $AXB$ at point $C$. The measure of this subtended angle remains invariant regardless of the precise location chosen for point $C$, provided $C$ remains on the circle and outside arc AXB$.\n\nTheorem 9 states that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle outside the arc.\n\nGiven a circle with centre C$, arc$AFB$, central angle$\angle BCA$subtended by arc$AFB$, and point$D$on the circle outside arc$AFB:\n\nTo show that \angle BCA = 2 \angle BDA:\n\nCase 1 (Extension of line segment DC$intersects the circle at point$E$inside arc$AFB):\n\nJoin D$to$C$and extend through$C$to point$E$. Triangle$\Delta DCB$is isosceles because$CB = CD = r$, giving$\angle CBD = \angle CDB$. By the exterior angle theorem applied to$\Delta DCB$, exterior angle$\angle BCE = \angle CBD + \angle CDB = 2 \angle BDC$. Triangle$\Delta ADC$is isosceles because$CA = CD = r$, giving$\angle CAD = \angle CDA$. By the exterior angle theorem applied to$\Delta ADC$, exterior angle$\angle ACE = \angle CAD + \angle CDA = 2 \angle CDA$. Summing both exterior angle equations yields$\angle BCA = \angle BCE + \angle ACE = 2 \angle BDC + 2 \angle CDA = 2 (\angle BDC + \angle CDA) = 2 \angle BDA$.

Case 2 (Extension of line segment $DC$ intersects the circle at point $E$ outside arc $AFB$):

By exterior angle properties of isosceles triangles $\Delta ADC$ and $\Delta BDC$, $\angle ACE = 2 \angle ADC$ and $\angle BCE = 2 \angle BDC$. Subtracting the smaller angle gives \angle ACB = \angle ACE - \angle BCE = 2 \angle ADC - 2 \angle BDC = 2 (\angle ADC - \angle BDC) = 2 \angle ADB$.\n\nAs a direct consequence of Theorem 9, all angles subtended by the same arc segment at points along the circle outside that arc are equal to one another. For arbitrary points E$,$F$, and$D$located on the circle outside arc$AXB$,$\angle AEB = \angle ADB = \angle AFB = \frac{1}{2} \angle ACB$.

When arc $AB$ is a semicircle, segment $AB$ is a diameter. The central angle subtended by a diameter is a straight angle measuring $180^\circ$. Consequently, the angle subtended by a diameter at any point $D$ on the circle is $\angle ADB = \frac{1}{2} (180^\circ) = 90^\circ$. Thus, the angle inscribed in a semicircle is always a right angle.

Subtended angle behavior varies relative to point positioning: for points strictly interior to the circle, subtended angles are strictly greater than inscribed circle angles; for points strictly exterior to the circle, subtended angles are strictly smaller; for all points lying exactly on the circular boundary outside the target arc, subtended angles are identical.

In a circle with centre $O$ and radius $12\,\text{cm}$, if central angle $\angle AOB = 60^\circ$, the length of chord $AB$ is $12\,\text{cm}$. Because $OA = OB = 12\,\text{cm}$ and $\angle AOB = 60^\circ$, triangle $\Delta AOB$ is equilateral, forcing chord $AB = 12\,\text{cm}$.

Concyclicity and Cyclic Quadrilaterals

Points that lie on the circumference of the same circle are defined as concyclic.

Theorem 10 states that if a line segment $AB$ joining two points $A$ and $B$ subtends equal angles at two other points $C$ and $D$ that lie on the same side of $AB$, then the four points $A$, $B$, $C$, and $D$ lie on a circle (they are concyclic).

Given line segment $AB$ and points $C, D$ on the same side of $AB$ with $\angle ACB = \angle ADB$:

To show that $A$, $B$, $C$, and $D$ are concyclic:

Proof by Contradiction: Non-collinear points $A$, $B$, and $C$ define a unique circle (Theorem 1). Assume $D$ does not lie on this circle:

If $D$ lies outside the circle, line segment $AD$ intersects the circle at some point $E$. Quadrilateral vertices $A$, $B$, $C$, and $E$ lie on the circle, making $\angle AEB = \angle ACB$ (angles subtended in the same arc segment). However, $\angle AEB$ is an exterior angle to triangle $\Delta BED$, requiring $\angle AEB > \angle ADB$. Substituting equality $\angle ACB = \angle ADB$ produces $\angle ACB > \angle ACB$, an impossibility.

If $D$ lies inside the circle, extending segment $AD$ meets the circle at point $F$. An exterior angle contradiction occurs similarly within \Delta BED$.\n\nBecause D$can neither lie outside nor inside the circle, point$D$must lie directly on the circle passing through$A$,$B$, and$C$. Hence,$A$,$B$,$C$, and$D are concyclic.\n\nWhen all four vertices of a quadrilateral (4-gon) are concyclic, the figure is called a cyclic quadrilateral.\n\nTheorem 11 states that the sum of two opposite angles of a cyclic quadrilateral is 180^\circ (opposite angles are supplementary).\n\nGiven cyclic quadrilateral ABCD$inscribed in a circle with centre$O:\n\nTo show that \angle BAD + \angle BCD = 180^\circ:\n\nFor arc BCD$, point$A$lies outside the arc. By Theorem 9, angle$\angle BAD$equals half the central angle subtended by arc$BCD$, which corresponds to$\text{reflex } \angle BOD$. Thus,$\angle BAD = \frac{1}{2} (\text{reflex } \angle BOD)$. For arc$BAD$, point$C$lies outside the arc. By Theorem 9,$\angle BCD$equals half the non-reflex central angle$\angle BOD$. Thus,$\angle BCD = \frac{1}{2} (\angle BOD)$.

Summing these terms yields $\angle BAD + \angle BCD = \frac{1}{2} (\text{reflex } \angle BOD + \angle BOD)$. A complete central rotation equals $360^\circ$. Therefore, $\angle BAD + \angle BCD = \frac{1}{2} (360^\circ) = 180^\circ$. By identical reasoning, \angle ABC + \angle ADC = 180^\circ$.\n\nA quadrilateral having interior angles \angle A = 80^\circ$,$\angle B = 110^\circ$,$\angle C = 100^\circ$, and$\angle D = 70^\circ$is cyclic because its opposite angles sum to$180^\circ$($\angle A + \angle C = 80^\circ + 100^\circ = 180^\circ$and$\angle B + \angle D = 110^\circ + 70^\circ = 180^\circ).\n\nTheorem 12 represents the converse of Theorem 11 and states that if two opposite angles of a quadrilateral add up to 180^\circ, then the vertices of the quadrilateral lie on a circle (they are concyclic).\n\nGiven quadrilateral ABCD$with$\angle BAD + \angle BCD = 180^\circ:\n\nTo show that ABCD is cyclic:\n\nProof by Contradiction: A unique circle passes through non-collinear points A$,$D$, and$B$. Assume point$C$does not lie on this circle and falls outside it. Let line segment$CD$intersect the circle at point$E$. Quadrilateral$ABED$is cyclic by construction. By Theorem 11,$\angle BAD + \angle BED = 180^\circ$. Given that$\angle BAD + \angle BCD = 180^\circ$, it follows that$\angle BCD = \angle BED$. However,$\angle BED$is an exterior angle of triangle$\Delta BEC$, making$\angle BED > \angle BCE = \angle BCD$. This contradiction refutes the assumption, proving$C$must lie on the circle. Thus, quadrilateral$ABCD$$ is cyclic.