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Circles - Angle Subtended by a Chord at a Point

Grade 9CBSE

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

🔑Concepts

Chord and Subtended Angle: A chord is a line segment joining any two points on the circumference of a circle. When the endpoints of a chord ABAB are joined to a point PP, the angle APB\angle APB is called the angle subtended by the chord at point PP. Visually, if PP is the center OO, the chord forms the base of an isosceles triangle AOB\triangle AOB.

Equal Chords and Center Angles: Equal chords of a circle (or of congruent circles) subtend equal angles at the center. If two chords ABAB and CDCD are equal in length (AB=CDAB = CD), then the angles they form at the center OO are equal, meaning AOB=COD\angle AOB = \angle COD.

Converse of Equal Chords Theorem: If the angles subtended by two chords of a circle at the center are equal, then the chords themselves must be equal in length. This means if AOB=COD\angle AOB = \angle COD, then AB=CDAB = CD.

Perpendicular from Center: The perpendicular line drawn from the center of a circle to a chord bisects the chord. Visually, if you draw a line from center OO meeting chord ABAB at MM such that OMA=90\angle OMA = 90^{\circ}, then MM is the midpoint of ABAB, so AM=MBAM = MB.

Equidistant Chords: Equal chords of a circle are equidistant from the center. The 'distance' of a chord from the center is the length of the perpendicular segment from the center to the chord. Visually, if AB=CDAB = CD, then the perpendicular distances OXOX and OYOY from the center OO to these chords will be equal (OX=OYOX = OY).

Chord Length and Distance Relationship: As the length of a chord increases, its distance from the center decreases. The longest chord of a circle is the diameter, and its distance from the center is 00.

Angle at the Circumference: For a fixed chord, the angle it subtends at the center is double the angle it subtends at any point on the remaining part of the circle. Visually, if a chord ABAB subtends AOB\angle AOB at center OO and ACB\angle ACB at a point CC on the major arc, then AOB=2ACB\angle AOB = 2\angle ACB.

📐Formulae

If Chord AB=Chord CD    AOB=CODChord\ AB = Chord\ CD \implies \angle AOB = \angle COD

If AOB=COD    Chord AB=Chord CD\angle AOB = \angle COD \implies Chord\ AB = Chord\ CD

Relationship between Radius (rr), Chord length (LL), and Distance from center (dd): r2=d2+(L2)2r^2 = d^2 + (\frac{L}{2})^2

Distance of chord from center: d=r2(L2)2d = \sqrt{r^2 - (\frac{L}{2})^2}

Angle relationship: Center=2×Circumference\angle Center = 2 \times \angle Circumference (for the same arc/chord)

💡Examples

Problem 1:

In a circle with center OO, two chords ABAB and CDCD are equal. If AOB=80\angle AOB = 80^{\circ}, find the value of COD\angle COD.

Solution:

  1. We are given that chord AB=CDAB = CD.
  2. According to the theorem, equal chords of a circle subtend equal angles at the center.
  3. Therefore, COD=AOB\angle COD = \angle AOB.
  4. Since AOB=80\angle AOB = 80^{\circ}, then COD=80\angle COD = 80^{\circ}.

Explanation:

This problem uses the direct application of the theorem stating that equal chords result in equal subtended angles at the center.

Problem 2:

A chord of length 16 cm16\text{ cm} is at a distance of 6 cm6\text{ cm} from the center of a circle. Find the radius of the circle.

Solution:

  1. Let the chord be AB=16 cmAB = 16\text{ cm} and the center be OO.
  2. Draw a perpendicular OMOM from OO to ABAB. Here, OM=6 cmOM = 6\text{ cm}.
  3. The perpendicular from the center bisects the chord, so AM=12×AB=162=8 cmAM = \frac{1}{2} \times AB = \frac{16}{2} = 8\text{ cm}.
  4. In the right-angled triangle OMA\triangle OMA, use Pythagoras theorem: OA2=OM2+AM2OA^2 = OM^2 + AM^2.
  5. OA2=62+82=36+64=100OA^2 = 6^2 + 8^2 = 36 + 64 = 100.
  6. OA=100=10 cmOA = \sqrt{100} = 10\text{ cm}. The radius is 10 cm10\text{ cm}.

Explanation:

This solution applies the property that a perpendicular from the center bisects the chord, forming a right-angled triangle where the radius is the hypotenuse.