Grade 9 Geometry: Properties and Theorems of Circles
Historical Foundations and Natural Patterns of Circles
Historical Observation:
Ancient human fascination with geometric forms is evident in early art and cave paintings.
In early cave paintings, the sun was frequently depicted as a circular shape.
In the prehistoric cave paintings of Gudahandi in Odisha, India, various geometric patterns are preserved, including triangles, squares, circles, and ovals.
These geometric shapes were inspired directly by natural phenomena observed in the human environment.
Circular Forms in Nature:
Concentric circles form on the surface of water when raindrops fall onto a still body of water.
Cross-sectional structures of plant stems display concentric circular boundaries.
The inflorescence of sunflowers demonstrates circular geometric arrangements.
Celestial bodies such as the full moon and the sun (notably during a total solar eclipse) present distinct circular profiles.
Mathematical Definitions and Fundamental Terminology
Two-Dimensional Plane Assumption:
Mathematical figures such as circles, triangles, and squares are defined assuming they lie on a flat, two-dimensional plane.
Formal Definition of a Circle:
A circle is defined as the set of all points in a plane that are equidistant from a given fixed point located on that same plane.
Concept of Locus:
A locus is defined as the set of all points that satisfy a specific geometric condition or rule.
Alternatively, a circle is the locus of points in a plane that remain at a constant distance from a fixed point.
Essential Elements of a Circle:
Centre: The fixed point on the plane from which all points on the circular boundary are at an equal distance.
Radius (): The fixed distance from the centre to any point lying on the circle.
Chord: A line segment connecting any two distinct points lying on the circle.
Diameter: A chord that passes directly through the centre of the circle. The length of the diameter is equal to twice the radius (). It represents the longest possible chord in any circle.
Subtended Angle: The angle formed at a point (such as the centre) by drawing straight line segments from that point to the two endpoints of a chord or arc. For a chord and centre , the angle subtended at the centre is .
Symmetries of a Circle and Geometric Figures
Rotational Symmetry of a Circle:
A circle possesses complete (infinite) rotational symmetry.
Rotating a circle about its centre through any angle leaves its visual appearance completely unchanged.
Physical metaphor: A rotating wheel appears identical at all times during rotation because every point on the boundary remains at distance from the centre.
Reflection Symmetry of a Circle:
Every line passing through the centre of a circle (i.e., every line containing a diameter) serves as an axis of reflection symmetry.
Folding a circular paper cut-out so that its curved boundaries overlap perfectly produces a crease line that represents a line of reflection symmetry, which always passes through the centre along a diameter.
Since infinitely many diameters exist, a circle has infinitely many lines of reflection symmetry.
Comparison with Regular Polygons:
Square: Has lines of reflection symmetry (two angle bisectors across diagonals and two mid-segment bisectors) and rotational symmetry of order (at angles of , , , and ).
Regular Pentagon: Has lines of reflection symmetry (lines joining each vertex to the midpoint of the opposite side) and rotational symmetry of order (at angles of , , , , and ).
Regular Hexagon: Has lines of reflection symmetry (three joining opposite vertices and three joining midpoints of opposite sides) and rotational symmetry of order (at angles of , , , , , and ).
Extremal Chord Lengths:
In a circle of radius , the length of the longest chord is the diameter, calculated as .
There is no non-zero smallest chord, because the length of a chord can be made arbitrarily close to as the chord segment approaches tangency at the circle's boundary.
Fundamental Locus Theorem for Two Points:
The locus of points equidistant from two given distinct points and is the perpendicular bisector of the line segment .
Every point lying on the perpendicular bisector of is equidistant from and , and every point equidistant from and lies on this perpendicular bisector.
Constructing Circles Through Given Points and Circumcircles
Circles Through Two Points:
Infinitely many circles can be drawn passing through two given points and .
For any circle passing through and , its centre satisfies .
Therefore, the centres of all possible circles passing through and must lie on the perpendicular bisector of segment .
Smallest and Largest Circles Through Two Points:
The smallest circle passing through and has its centre at the midpoint of segment , making segment its diameter.
The radius of this smallest circle is .
There is no upper bound on the radius of a circle passing through and ; as the centre moves further away along the perpendicular bisector, the radius approaches infinity (), and the curvature of the circle approaches zero (appearing increasingly flat like a straight line).
Circles Through Three Points:
Collinear Points Case: No circle can pass through three collinear points , , and C$.\n - *Explanation*: If ABCABBCOOA = OB = OC. Consequently, no straight line can intersect a circle in more than two distinct points.\n - **Non-Collinear Points Case (Theorem 1)**:\n - **Theorem 1**: *There is a unique circle passing through three non-collinear points.*\n - *Proof*: Let ABCABACABCOOABOA = OBOACOA = OCOA = OB = OCOOAABC$.
Circumcircle and Circumcentre of a Triangle:
The unique circle passing through the three vertices , , and of is called the circumcircle of .
The circumcircle is said to circumscribe the triangle, and the triangle is inscribed in the circle.
The centre of this circle is called the circumcentre of , defined as the point of concurrence of the perpendicular bisectors of the triangle's three sides.
Position of Circumcentre depending on Triangle Geometry:
Acute-Angled Triangle: The circumcentre lies strictly inside the triangle.
- **Obtuse-Angled Triangle**: The circumcentre lies strictly **outside** the triangle.
- **Right-Angled Triangle**: The circumcentre lies exactly at the **midpoint of the hypotenuse**.
Theorems on Chords and Subtended Angles at the Centre
Subtended Angle Properties:
When a chord of fixed length is rotated around the centre of a circle, the angle it subtends at the centre remains invariant.
Theorem 2 (Equal Chords Subtend Equal Angles):
Theorem 2: Equal chords of a circle subtend equal angles at the centre of the circle.
Given: A circle with centre and two chords and such that AB = DE$.\n - *To Prove*: \angle ACB = \angle DCE.\n - *Proof*:\n - In triangles \Delta CAB\Delta CDE:\n - CA = CD = r (radii of the same circle).\n - CB = CE = r (radii of the same circle).\n - AB = DE (given).\n - By Side-Side-Side (SSS\Delta CAB \cong \Delta CDE$.
Since corresponding parts of congruent triangles are equal (), \angle ACB = \angle DCE$.\n\n- Theorem 3 (Converse of Theorem 2):\n - **Theorem 3**: *Chords of a circle that subtend equal angles at the centre are equal in length.*\n - *Given*: A circle with centre CABDE\angle ACB = \angle DCE$.
To Prove: AB = ED$.\n - *Proof*:\n - In triangles \Delta ACB\Delta DCE:\n - AC = DC = r (radii of the circle).\n - BC = EC = r (radii of the circle).\n - \angle ACB = \angle DCE (given).\n - By Side-Angle-Side (SAS\Delta ACB \cong \Delta DCE$.
By , AB = ED$.\n\n# Midpoints, Perpendicular Bisectors, and Chord Distances\n\n- Theorem 4 (Centre to Midpoint is Perpendicular):\n - **Theorem 4**: *The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.*\n\n\n\n - *Given*: A circle with centre CABMAM = BM$.
To Prove: (i.e., ).
Proof:
Consider and :
(radii of the circle).
( is the midpoint of ).
(common side).
By congruence criterion, \Delta CMA \cong \Delta CMB$.\n - Therefore, \angle CMA = \angle CMBCPCT$.
Since forms a straight line, \angle CMA + \angle CMB = 180^\circ$.\n - Substituting \angle CMA = \angle CMB2\angle CMA = 180^\circ \implies \angle CMA = 90^\circ$.
Hence, CM \perp AB$.\n\n- Theorem 5 (Perpendicular Bisects the Chord):\n - **Theorem 5**: *The perpendicular from the centre of a circle to a chord of the circle bisects the chord.*\n - *Given*: A circle with centre CABCM \perp AB$.
To Prove: AM = BM$.\n - *Proof*:\n - In right-angled triangles \Delta CMA\Delta CMB:\n - \angle CMA = \angle CMB = 90^\circ (given).\n - Hypotenuse CA = \text{Hypotenuse } CB = r (radii).\n - Leg CM = \text{Leg } CM (common side).\n - By Right angle-Hypotenuse-Side (RHS\Delta CMA \cong \Delta CMB$.
By , . Thus, bisects AB$.\n\n- Theorem 6 (Equal Chords are Equidistant from Centre):\n - **Theorem 6**: *Chords of a circle having the same length are all at the same distance from the centre of the circle.*\n\n\n\n - *Given*: A circle with centre CAB = FGCE \perp ABCH \perp FG$.
To Prove: CE = CH$.\n - *Proof Method 1 (Triangle Congruence and Altitudes)*:\n - In \Delta CAB\Delta CFGCA = CF = rCB = CG = rAB = FG$.
By congruence, \Delta CAB \cong \Delta CFG$.\n - Since corresponding altitudes of congruent triangles are equal, CE = CH$.
Proof Method 2 (RHS Congruence):
Since perpendiculars from the centre bisect the chords, and FH = \frac{1}{2}FG$.\n - Given AB = FGAE = FH$.
In right triangles and :
\angle CEA = \angle CHF = 90^\circ$.\n - Hypotenuse CA = \text{Hypotenuse } CF = r$.
Leg AE = \text{Leg } FH$.\n - By RHS\Delta CEA \cong \Delta CHF$.
By , CE = CH$.\n - *Proof Method 3 (Baudhāyana–Pythagoras Theorem)*:\n - In right \Delta CEACA^2 = CE^2 + AE^2 \implies r^2 = CE^2 + \left(\frac{AB}{2}\right)^2$.
In right , CF^2 = CH^2 + FH^2 \implies r^2 = CH^2 + \left(\frac{FG}{2}\right)^2$.\n - Equating expressions for r^2CE^2 + \left(\frac{AB}{2}\right)^2 = CH^2 + \left(\frac{FG}{2}\right)^2$.
Since , \left(\frac{AB}{2}\right)^2 = \left(\frac{FG}{2}\right)^2$.\n - Subtracting this term from both sides yields CE^2 = CH^2 \implies CE = CH$.
Theorem 7 (Converse of Theorem 6):
Theorem 7: Chords of a circle that are equidistant from the centre have equal length.
Given: , , and CE = CH$.\n - *To Prove*: AB = FG$.
Proof:
In right triangles and , hypotenuses , legs CE = CH$.\n - By RHS\Delta CEA \cong \Delta CHF \implies AE = FH$.
Since and , it follows that AB = FG$.\n\n# Distance and Relative Lengths of Unequal Chords\n\n- Theorem 8 (Unequal Chords and Distance from Centre):\n - **Theorem 8**: *Let ABDECAB > DECABCDE$.*
Given: Circle with centre ; chords ; perpendiculars and CG \perp DE$.\n - *To Prove*: CF < CG$.
Proof:
and are midpoints of and respectively (, ).
Since , it follows that AF > GD \implies AF^2 > GD^2$.\n - In right triangle \Delta CFAAC^2 = CF^2 + AF^2$.
In right triangle : CD^2 = CG^2 + GD^2$.\n - Since AC = CD = r (radii):\n CF^2 + AF^2 = CG^2 + GD^2\n CF^2 - CG^2 = GD^2 - AF^2\n - Because AF^2 > GD^2GD^2 - AF^2 < 0$.
Therefore, CF^2 - CG^2 < 0 \implies CF^2 < CG^2 \implies CF < CG$.\n- Fundamental Consequences:\n - The chord closest to the centre is a diameter, whose perpendicular distance from the centre is 0\,cm2r).\n - As a chord is moved further away from the centre, its length strictly decreases.\n - At a perpendicular distance d = r0\,cm (representing a point of tangency on the circle).\n\n# Arcs, Central Angles, and Angles Subtended on the Circle\n\n- Definitions of Arcs and Angles:\n - **Arc**: A connected portion of a circle defined by two endpoints lying on the circle.\n - **Minor Arc**: The shorter arc connecting two points AB180^\circ$.
Major Arc: The longer arc connecting two points and . Subtends a central angle greater than 180^\circ$.\n - **Central Angle**: The angle swept by radii moving from one endpoint of an arc to the other endpoint along the arc.\n\n- Theorem 9 (Degree Measure Theorem / Angle at Centre):\n - **Theorem 9**: *The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.*\n - *Given*: An arc AFBC\angle ACBDAFB$.
To Prove: \angle BCA = 2\angle BDA$.\n - *Proof Case 1 (Line DCAFBE)*:\n - Join DCEAFB$.
In isosceles (), base angles \angle CBD = \angle CDB$.\n - By the Exterior Angle Theorem for \Delta BCD:\n \angle BCE = \angle CBD + \angle CDB = 2\angle BDC\n - In isosceles \Delta ADCCA = CD = r\angle CAD = \angle CDA$.
By the Exterior Angle Theorem for :
Adding the two equations:
Proof Case 2 (Line extended cuts the circle at outside arc ):
and \angle BCE = 2\angle BDC$.\n - Subtracting the two equations:\n \angle ACB = \angle ACE - \angle BCE = 2\angle ADC - 2\angle BDC = 2(\angle ADC - \angle BDC) = 2\angle ADB\n\n- Invariance of Subtended Angles (Angles in the Same Segment):\n - For any points DEFAXB:\n \angle ADB = \angle AEB = \angle AFB = \frac{1}{2}\angle ACB\n - All angles subtended by a given arc at any points on the remaining part of the circle are equal to each other.\n\n- Corollaries of Theorem 9:\n - **Corollary 1 (Angle in a Semicircle)**: *The angle subtended by a diameter at any point on the circle is 90^\circ.*\n - *Proof*: A diameter subtends a central angle of 180^\circD\frac{1}{2} \times 180^\circ = 90^\circ$.
Spatial Angle Relations:
For points strictly inside the circle, the subtended angle is strictly greater than the angle on the circle boundary ().
For points strictly outside the circle, the subtended angle is strictly less than the angle on the circle boundary ().
Concyclicity and Properties of Cyclic Quadrilaterals
Concyclic Points:
A set of points is defined as concyclic if all of them lie on the boundary of a single, common circle.
Theorem 10 (Concyclicity Criterion):
Theorem 10: If a line segment joining two points subtends equal angles at two other points that lie on the same side of , then the four points lie on a circle (i.e., they are concyclic).
Proof:
Three non-collinear points determine a unique circumcircle.
Suppose does not lie on this circle. Then lies either inside or outside the circle.
Let segment (or its extension) intersect the circle at point E$.\n - Since CEAB\angle AEB = \angle ACB$.
If were outside, would be an exterior angle to , implying . Since (given), this yields , a contradiction.
An analogous contradiction arises if were inside.
Hence, must lie on the circle, making concyclic.
Cyclic Quadrilateral:
A quadrilateral whose four vertices are concyclic is called a cyclic quadrilateral.
Theorem 11 (Opposite Angles of a Cyclic Quadrilateral):
Theorem 11: The sum of two opposite angles of a cyclic quadrilateral is (i.e., opposite angles are supplementary).
Given: Cyclic quadrilateral inscribed in a circle with centre O$.\n - *To Prove*: \angle BAD + \angle BCD = 180^\circ\angle ABC + \angle ADC = 180^\circ$.
Proof:
subtends arc ; it equals half the reflex central angle :
subtends arc ; it equals half the non-reflex central angle :
Summing the two angles:
Since a complete rotation about centre equals :
Similarly, \angle ABC + \angle ADC = 180^\circ$.\n\n- Theorem 12 (Converse of Theorem 11):\n - **Theorem 12**: *If two opposite angles of a quadrilateral add up to 180^\circ, then the vertices of the quadrilateral lie on a circle (they are concyclic).*\n - *Proof*:\n - Let ABCD\angle BAD + \angle BCD = 180^\circ$.
Draw the unique circumcircle through non-collinear vertices A, B, D$.\n - If CCDE$.
is a cyclic quadrilateral, so \angle BAD + \angle BED = 180^\circ$.\n - Since \angle BAD + \angle BCD = 180^\circ\angle BCD = \angle BED$.
But is an exterior angle of , so , causing a contradiction.
Thus, must lie on the circle, proving is cyclic.
Complete Solutions to End-of-Chapter Exercises and Proofs
Exercise Set 5.1 & Think/Reflect Questions:
Jamuna's paper circle centre location: Amina told Jamuna to fold the circular paper twice along two different crease lines (diameters). The point where the two crease lines intersect is the exact centre of the circle.
Least radius of a circle through two points : The least possible radius is r_{\text{min}} = \frac{1}{2}AB$.\n 3. *Three collinear points A, B, CPPA = PB = PCABBC are parallel lines that never intersect.\n\n- Exercise Set 5.3 Questions:\n 1. *Distance between parallel chords on opposite sides*:\n - Given: Radius r = 5\,cm6\,cm8\,cm$.
Half-lengths of chords: and \frac{8}{2} = 4\,cm$.\n - Distance of first chord from centre: d_1 = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\,cm$.
Distance of second chord from centre: d_2 = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\,cm$.\n - Total distance between midpoints (chords on opposite sides): d_1 + d_2 = 4\,cm + 3\,cm = 7\,cm$.
Exercise Set 5.5 & End-of-Chapter Problems:
Length of chord with and perpendicular distance :
Formula: \text{Length} = 2\sqrt{r^2 - d^2} = 2\sqrt{7^2 - 6^2} = 2\sqrt{49 - 36} = 2\sqrt{13}\,cm$.\n 2. *General formula justification*:\n - In right triangle formed by radius rd\frac{L}{2}:\n r^2 = d^2 + \left(\frac{L}{2}\right)^2 \implies \left(\frac{L}{2}\right)^2 = r^2 - d^2 \implies \frac{L}{2} = \sqrt{r^2 - d^2} \implies L = 2\sqrt{r^2 - d^2}\n 3. *End-of-Chapter Question 1*: Chord distance d = 5\,cmr = 13\,cm$.
\text{Length} = 2\sqrt{13^2 - 5^2} = 2\sqrt{169 - 25} = 2\sqrt{144} = 2 \times 12 = 24\,cm$.\n 4. *End-of-Chapter Question 3*: Diameter = 26\,cm \implies r = 13\,cm= 24\,cm$.
Perpendicular distance d = \sqrt{13^2 - \left(\frac{24}{2}\right)^2} = \sqrt{169 - 144} = \sqrt{25} = 5\,cm$.\n 5. *End-of-Chapter Question 4*: Radius r = 15\,cmd = 9\,cm$.
\text{Chord Length} = 2\sqrt{15^2 - 9^2} = 2\sqrt{225 - 81} = 2\sqrt{144} = 2 \times 12 = 24\,cm$.\n 6. *End-of-Chapter Question 7*: Cyclic quadrilateral ABCD\angle A = 75^\circ\angle B = 110^\circ$.
\angle C = 180^\circ - \angle A = 180^\circ - 75^\circ = 105^\circ$.\n - \angle D = 180^\circ - \angle B = 180^\circ - 110^\circ = 70^\circ$.
End-of-Chapter Question 8: Cyclic quadrilateral with and \angle R = (3x - 20)^\circ$.\n - Opposite angles sum to 180^\circ:\n (2x + 10) + (3x - 20) = 180 \implies 5x - 10 = 180 \implies 5x = 190 \implies x = 38\n - \angle P = 2(38) + 10 = 76 + 10 = 86^\circ$.
\angle R = 3(38) - 20 = 114 - 20 = 94^\circ$.\n 8. *End-of-Chapter Question 9*: Chord length 16\,cm6\,cm$.
Half-length = 8\,cm$.\n - Radius r = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,cm$.
End-of-Chapter Question 10: Area of cyclic quadrilateral with sides .
This is a kite formed by two right triangles sharing a diagonal (the diameter).
Diagonal dividing into two isosceles pairs or right triangles: hypotenuse d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ units}$.\n - Total Area = 2 \times \left(\frac{1}{2} \times 5 \times 12\right) = 60\text{ square units}$.
End-of-Chapter Question 14: Parallelogram inscribed in a circle must be a rectangle.
Proof: In parallelogram , opposite angles are equal: and \angle B = \angle D$.\n - Since ABCD\angle A + \angle C = 180^\circ \implies 2\angle A = 180^\circ \implies \angle A = 90^\circ$.
A parallelogram with a right angle is a rectangle.
End-of-Chapter Question 16: Shape formed by midpoints of all chords of fixed length.
Proof: Chords of equal length are all equidistant from the centre (distance ).
The locus of midpoints at a constant distance from centre is a concentric circle of radius d$.\n 12. *End-of-Chapter Question 18*: Parallel chords 10\,cm24\,cm7\,cm$.
Let radius be , distance of chord from centre be x$.\n - Half-lengths: 12\,cm5\,cm$.
For chord : r^2 = x^2 + 12^2 = x^2 + 144$.\n - For chord 10\,cmr^2 = (x + 7)^2 + 5^2 = (x + 7)^2 + 25$.
Equating: x^2 + 144 = x^2 + 14x + 49 + 25 \implies x^2 + 144 = x^2 + 14x + 74$.\n - 14x = 144 - 74 = 70 \implies x = 5\,cm$.
Radius r = \sqrt{5^2 + 144} = \sqrt{25 + 144} = \sqrt{169} = 13\,cm$.\n 13. *End-of-Chapter Question 19*: Regular hexagon inscribed in circle of radius r$.
Side length of regular hexagon (subtends central angle, forming equilateral triangle with radii).
Perpendicular distance from centre to each side d = \sqrt{r^2 - (r/2)^2} = \sqrt{\frac{3r^2}{4}} = \frac{r\sqrt{3}}{2}$.\n 14. *End-of-Chapter Question 21*: Exterior angle of cyclic quadrilateral equal to interior opposite angle.\n - Let ECD$.
Linear pair: \angle ADC + \angle CDE = 180^\circ \implies \angle CDE = 180^\circ - \angle ADC$.\n - Cyclic opposite angles: \angle ABC + \angle ADC = 180^\circ \implies \angle ABC = 180^\circ - \angle ADC$.
Therefore, \angle CDE = \angle ABC$.\n\n# Comprehensive Chapter Summary\n\n- Definition & Locus:\n - A circle is the set (locus) of all points in a plane at a fixed radius rO$.
Symmetry Rules:
Reflection symmetry across any line containing a diameter (infinitely many lines).
Complete rotational symmetry about its centre through any angle \theta$.\n- Point-Circle Incidence:\n - Infinitely many circles pass through two given points; their centres lie on the perpendicular bisector of the line segment connecting the points.\n - Exactly one unique circle (circumcircle) passes through three non-collinear points; its centre (circumcentre) is the intersection of the perpendicular bisectors of the sides.\n - Zero circles pass through three collinear points.\n- Chord Properties:\n - Equal chords subtend equal angles at the centre, and conversely, chords subtending equal central angles are equal in length.\n - The line segment connecting the centre to the midpoint of a chord is perpendicular to the chord; conversely, a perpendicular from the centre to a chord bisects the chord.\n - Equal chords are equidistant from the centre; conversely, chords equidistant from the centre are equal in length.\n - For unequal chords, the longer chord is closer to the centre, with the diameter being the longest chord (distance 0\,cm).\n- Arc and Angle Rules:\n - The central angle subtended by an arc is double the angle subtended by the same arc at any point on the remaining circumference.\n - Angles subtended by an arc at any points in the same segment of a circle are equal.\n - The angle subtended by a diameter at any point on the circle is a right angle (90^\circ).\n- Concyclicity & Cyclic Quadrilaterals:\n - If a segment subtends equal angles at two points on the same side, all four points are concyclic.\n - The opposite angles of a cyclic quadrilateral are supplementary (\text{sum} = 180^\circ$$).
Conversely, if a quadrilateral has supplementary opposite angles, it is a cyclic quadrilateral.