Grade 11 Mathematics Study Notes: Euclidean Geometry (Circle Geometry)

Centre of a Circle: Defined as a point inside the circle that is equidistant from all the points on the circumference of the circle.
Radius: A line from the centre to any point on the circumference of the circle.
Diameter: A line segment that passes through the centre with its endpoints located on the circumference of the circle.
Secant: A straight line that intersects a circle at exactly two points.
Chord: A line segment with both endpoints on the circumference of the circle.
Segment: A region of a circle bounded by a chord and a corresponding arc lying between the chord's endpoints.
Arc: A smooth curve joining two endpoints on the circumference.
Sector: A pie-shaped part of a circle made of an arc along with its two radii.
Tangent: A line that touches the circle at a single unique point.
Cyclic Quadrilateral: A quadrilateral which has all four of its vertices lying on the circumference of a circle.

Theorem Statement: A line drawn from the centre of a circle perpendicular to a chord bisects the chord.
Mathematical Representation: If $OM \perp KL$ , then $KM = ML$ (Reason: $\text{line from centre } \perp \text{ to the chord}$ ).
Theorem 1 Converse: A line drawn from the centre to the midpoint of a chord is perpendicular to the chord. If $KM = ML$ , then $OM \perp KL$ (Reason: $\text{line from centre to midpt of the chord }$ ).
Activity Examples: * Finding Length: Given center $O$ , $MO = 3\text{ units}$ , and chord $KL = 8\text{ units}$ . * Since $MO \perp KL$ , $KM = \frac{1}{2} KL = 4\text{ units}$ . * Using the Theorem of Pythagoras in $\triangle KOM$ : $KO^2 = OM^2 + KM^2$ . * $KO^2 = (3)^2 + (4)^2 = 9 + 16 = 25$ . * $KO = 5\text{ units}$ . * Algebraic Application: Given circle center $O$ , chord $JL = 24\text{ units}$ , $HJ = 12\sqrt{5}$ , distance $MG = 6\text{ units}$ , and $OM = x$ . * $HM = x + 6 + x = 2x + 6$ . * In $\triangle HMJ$ , $HM^2 + JM^2 = JH^2$ (Pythagoras). * $(2x + 6)^2 + 12^2 = (12\sqrt{5})^2$ . * $x^2 + 6x - 135 = 0$ . * Factors: $(x - 9)(x + 15) = 0$ . Thus, $x = 9$ (as length cannot be negative). * $\text{Radius} = x + 6 = 9 + 6 = 15$ .

Theorem Statement: The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circumference (on the same side of the chord as the centre).
Mathematical Representation: $\angle \text{at centre} = 2 \times \angle \text{at circumference}$ .
Examples: * If the angle at the circumference is $40^{\circ}$ , the angle at the centre is $80^{\circ}$ . * If the angle at the centre is $140^{\circ}$ , the angle at the circumference is $70^{\circ}$ .
Important Connection: Radii of the same circle are equal. This often forms isosceles triangles (e.g., $\triangle OBC$ where $OB = OC$ ), meaning angles opposite equal sides are equal.

Theorem Statement: The angle subtended by the diameter at the circumference of the circle is $90^{\circ}$ .
Reasoning: This is a special case of Theorem 2, where the angle at the centre is a straight line ( $180^{\circ}$ ).
Acceptable Reason: $\angle \text{s in semi-circle}$ .
Converse of Theorem 3: If the angle subtended by a chord at the circumference is $90^{\circ}$ , then that chord is a diameter. * Application: To prove $BK$ is a diameter, show $\angle BAK = 90^{\circ}$ , then conclude $BK$ is a diameter (Reason: $\text{converse } \angle \text{s in semi-circle}$ ).

Theorem Statement: Angles subtended by a chord or arc of the circle, on the same side of the chord, are equal.
Reasoning: $\angle \text{s in the same segment}$ .
Application: If $\angle LAK$ and $\angle LBK$ are subtended by arc $LK$ , then $\angle LAK = \angle LBK$ .
Algebraic Activity: Given $\angle J = 3x + 10^{\circ}$ and $\angle K = 2x + 80^{\circ}$ subtended by the same arc, then: * $3x + 10^{\circ} = 2x + 80^{\circ}$ * $x = 70^{\circ}$ .
Theorem 4 Corollaries: * Equal chords subtend equal angles at the circumference. * Equal chords subtend equal angles at the centre. * Equal chords in equal circles subtend equal angles at the circumference.

Theorem Statement: The opposite angles of a cyclic quadrilateral are supplementary (add up to $180^{\circ}$ ).
Mathematical Representation: $\hat{K} + \hat{N} = 180^{\circ}$ and $\hat{L} + \hat{M} = 180^{\circ}$ .
Reasoning: $\text{Opp } \angle \text{s of cyclic quad}$ .
Proving a Quadrilateral is Cyclic: If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic (Converse Theorem 5).

Theorem Statement: The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
Mathematical Representation: If side $MN$ is produced to $P$ , then $\angle LNP = \hat{K}$ .
Reasoning: $\text{Ext } \angle \text{ of cyclic quad}$ .
Proving a Quadrilateral is Cyclic: * Show the exterior angle equals the interior opposite angle (Reason: $\text{converse ext } \angle \text{ of cyclic quad}$ ). * Show a line segment joining two points subtends equal angles at two other points on the same side of the line (Reason: $\text{line subtends equal } \angle \text{s}$ ).

Theorem Statement: The tangent to a circle is perpendicular to the radius (or diameter) at the point of contact.
Mathematical Representation: Radius $OM \perp$ Tangent $LMN$ results in $\angle OMN = 90^{\circ}$ .
Reasoning: $\tan \perp \text{radius}$ or $\tan \perp \text{diameter}$ .
How to prove a line is a tangent: Show that the line is perpendicular to the radius at the point of contact (Reason: $\text{converse tan } \perp \text{radius}$ ).

Theorem Statement: The angle between the tangent to a circle and a chord drawn from the point of contact is equal to the angle in the alternate segment.
Mathematical Representation: If $PNM$ is a tangent and $NK$ is a chord, then $\angle KNM = \angle NLK$ .
Reasoning: $\text{tan chord theorem}$ .
How to prove a line is a tangent: If a line is drawn through the endpoint of a chord making an angle equal to the angle in the alternate segment, the line is a tangent (Reason: $\text{converse tan chord theorem}$ ).

Theorem Statement: Two tangents drawn to a circle from the same point outside the circle are equal in length.
Mathematical Representation: If $NM$ and $QM$ are tangents from point $M$ , then $NM = QM$ .
Reasoning: $\text{tans from same pt}$ .
Consequence: Since the two tangents are equal, the triangle formed with the chord of contact is an isosceles triangle ( $\triangle NMQ$ is isosceles), meaning $\angle MNQ = \angle MQN$ .