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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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The perpendicular from the center of a circle to a chord bisects the chord. Conversely, the line joining the center of a circle to the midpoint of a chord is perpendicular to the chord. In the diagram, if OM⊥ABOM \perp AB, then AM=MBAM = MB.

A circle showing center O and a perpendicular line OM bisecting chord AB at M.
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Equal chords of a circle are equidistant from the center. This means if two chords have the same length, their perpendicular distances from the center are equal.

Two equal parallel chords at equal distances d1 and d2 from the center.
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The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. If ∠AOB\angle AOB is the central angle, then ∠AOB=2∠ACB\angle AOB = 2 \angle ACB.

Angle subtended at the center O is twice the angle at circumference point C.
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Angles in the same segment of a circle are equal. This implies that any two angles subtended by the same arc at the circumference are identical.

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The angle in a semi-circle is a right angle (90∘90^\circ). Any triangle formed using the diameter as one side and a third point on the circumference is a right-angled triangle.

📐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).

Problem 3:

Two parallel chords of lengths 16 cm16 \text{ cm} and 12 cm12 \text{ cm} are on the same side of the center of a circle of radius 10 cm10 \text{ cm}. Find the distance between the two chords.

Two parallel chords AB and CD on the same side of center O.

Solution:

Let the center of the circle be OO and the chords be AB=16 cmAB = 16 \text{ cm} and CD=12 cmCD = 12 \text{ cm}. Let OM⊥ABOM \perp AB and ON⊥CDON \perp CD. MM and NN are midpoints, so AM=8 cmAM = 8 \text{ cm} and CN=6 cmCN = 6 \text{ cm}. In △OMA\triangle OMA: OM2=OA2−AM2OM^2 = OA^2 - AM^2 OM2=102−82=100−64=36OM^2 = 10^2 - 8^2 = 100 - 64 = 36 OM=6 cmOM = 6 \text{ cm} In △ONC\triangle ONC: ON2=OC2−CN2ON^2 = OC^2 - CN^2 ON2=102−62=100−36=64ON^2 = 10^2 - 6^2 = 100 - 36 = 64 ON=8 cmON = 8 \text{ cm} Distance between chords =ON−OM=8−6=2 cm= ON - OM = 8 - 6 = 2 \text{ cm}.

Explanation:

We use the Pythagorean theorem in two right-angled triangles formed by the radii and the perpendiculars to the chords. Since the chords are on the same side, we subtract the distances from the center.

Problem 4:

In a circle with center OO, the chord ABAB is equal to the radius of the circle. Find the angle subtended by this chord at a point on the major arc.

Equilateral triangle OAB inside a circle with angle at point C on circumference.

Solution:

In △OAB\triangle OAB, OA=OBOA = OB (radii) and AB=OAAB = OA (given). Therefore, △OAB\triangle OAB is an equilateral triangle. So, ∠AOB=60∘\angle AOB = 60^\circ. Let CC be a point on the major arc. By the Central Angle Theorem: ∠ACB=12×∠AOB\angle ACB = \frac{1}{2} \times \angle AOB ∠ACB=12×60∘=30∘\angle ACB = \frac{1}{2} \times 60^\circ = 30^\circ.

Explanation:

An equilateral triangle is formed when the chord length equals the radius. The angle at the center is 60∘60^\circ, and the angle at the circumference is half of that.