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Geometry - Circle (Chord properties and Arc properties)

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Perpendicular Bisector Property: A perpendicular line drawn from the center of a circle to a chord bisects the chord into two equal parts. Visually, in a circle with center OO and chord ABAB, if a line OMOM is drawn such that OM⊥ABOM \perp AB, then AM=MBAM = MB. This creates two right-angled triangles △OMA\triangle OMA and △OMB\triangle OMB where the radius OAOA or OBOB acts as the hypotenuse.

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Equal Chords and Distances: Chords that are equal in length are located at an equal distance from the center of the circle. Conversely, chords that are equidistant from the center are equal in length. If you visualize two chords ABAB and CDCD in a circle with center OO, and the perpendicular distances OMOM and ONON are equal, then AB=CDAB = CD.

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Angles Subtended by Chords: Equal chords of a circle subtend equal angles at the center. If chord AB=chord CDAB = \text{chord } CD, then the angle formed at the center ∠AOB=∠COD\angle AOB = \angle COD. This property is fundamental for proving congruency between triangles formed by radii and chords.

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The Central Angle Theorem: The angle subtended by an arc at the center of a circle is double the angle subtended by the same arc at any point on the remaining part of the circumference. Visually, if an arc PQPQ forms ∠POQ\angle POQ at center OO and ∠PAQ\angle PAQ at point AA on the circle, then ∠POQ=2∠PAQ\angle POQ = 2\angle PAQ.

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Angles in the Same Segment: All angles subtended by the same arc (or chord) in the same segment of a circle are equal. If you draw several triangles sharing the same base chord ABAB and having their third vertex on the same arc, all those vertex angles (e.g., ∠APB\angle APB, ∠AQB\angle AQB) will be identical in measure.

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Angle in a Semicircle: An angle subtended by the diameter at any point on the circumference of the circle is always a right angle (90∘90^\circ). If ABAB is a diameter passing through center OO, and CC is any point on the circle, then the triangle ABCABC formed will always have ∠ACB=90∘\angle ACB = 90^\circ.

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Congruent Arcs and Chords: In the same circle or congruent circles, if two arcs are equal in measure, then their corresponding chords are also equal. This means arc length and chord length are directly related; larger arcs correspond to larger chords up until the diameter.

📐Formulae

Relationship between Radius (rr), Chord length (cc), and Perpendicular distance (dd): r2=d2+(c2)2r^2 = d^2 + (\frac{c}{2})^2

Length of a chord: c=2r2−d2c = 2\sqrt{r^2 - d^2}

Distance of chord from center: d=r2−(c2)2d = \sqrt{r^2 - (\frac{c}{2})^2}

Central Angle Theorem: ∠center=2×∠circumference\angle \text{center} = 2 \times \angle \text{circumference}

Sum of angles in a triangle (often used with isosceles triangles formed by radii): ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ

💡Examples

Problem 1:

A chord of length 24 cm24 \text{ cm} is drawn in a circle of radius 13 cm13 \text{ cm}. Find the perpendicular distance of the chord from the center of the circle.

Solution:

  1. Let the chord be AB=24 cmAB = 24 \text{ cm} and the center be OO.
  2. Draw a perpendicular OMOM from OO to the chord ABAB. According to circle properties, OMOM bisects ABAB. Therefore, AM=12×AB=12×24=12 cmAM = \frac{1}{2} \times AB = \frac{1}{2} \times 24 = 12 \text{ cm}.
  3. In the right-angled triangle OMAOMA, the radius OA=13 cmOA = 13 \text{ cm} is the hypotenuse.
  4. Using Pythagoras theorem: OA2=OM2+AM2OA^2 = OM^2 + AM^2.
  5. 132=OM2+12213^2 = OM^2 + 12^2
  6. 169=OM2+144169 = OM^2 + 144
  7. OM2=169−144=25OM^2 = 169 - 144 = 25
  8. OM=25=5 cmOM = \sqrt{25} = 5 \text{ cm}.

Explanation:

This problem uses the property that a perpendicular from the center bisects the chord, allowing us to use the Pythagoras theorem on the resulting right-angled triangle.

Problem 2:

In a circle with center OO, an arc ABAB subtends an angle of 130∘130^\circ at the center. Find the measure of the angle subtended by the major arc ABAB at a point PP on the minor arc.

Solution:

  1. The angle subtended by the minor arc ABAB at the center is ∠AOB=130∘\angle AOB = 130^\circ.
  2. The reflex angle ∠AOB\angle AOB (representing the major arc) is 360∘−130∘=230∘360^\circ - 130^\circ = 230^\circ.
  3. According to the Central Angle Theorem, the angle subtended by an arc at the circumference is half the angle it subtends at the center.
  4. The angle at point PP on the minor arc is subtended by the major arc ABAB.
  5. Therefore, ∠APB=12×(Reflex ∠AOB)\angle APB = \frac{1}{2} \times (\text{Reflex } \angle AOB).
  6. ∠APB=12×230∘=115∘\angle APB = \frac{1}{2} \times 230^\circ = 115^\circ.

Explanation:

The Central Angle Theorem applies to both minor and major arcs. When finding the angle at the circumference facing the center, we must use the corresponding central angle (reflex angle for the major arc).