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Circles - Relate chords, subtended angles, and perpendicular bisector properties

Grade 9CBSE

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

🔑Concepts

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Equal chords of a circle subtend equal angles at the center. Conversely, if the angles subtended by the chords of a circle at the center are equal, then the chords are equal.

A circle showing two equal chords AB and CD subtending equal angles at center O.
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The perpendicular from the center of a circle to a chord bisects the chord. This means if OM⊥ABOM \perp AB, then AM=MB=12ABAM = MB = \frac{1}{2}AB.

A perpendicular from center O to chord AB meeting at midpoint M.
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Equal chords of a circle (or of congruent circles) are equidistant from the center. Conversely, chords equidistant from the center of a circle are equal in length.

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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.

📐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.

Problem 3:

In a circle of radius 5 cm5\text{ cm}, ABAB is a chord such that AB=8 cmAB = 8\text{ cm}. Calculate the distance of the chord from the center OO.

Triangle OMA with OA=5, AM=4 and OM as the perpendicular distance.

Solution:

  1. Let OMOM be the perpendicular from center OO to chord ABAB. By the perpendicular bisector theorem, MM is the midpoint of ABAB.
  2. AM=AB2=82=4 cmAM = \frac{AB}{2} = \frac{8}{2} = 4\text{ cm}.
  3. In right-angled triangle △OMA\triangle OMA, by Pythagoras theorem: OA2=OM2+AM2OA^2 = OM^2 + AM^2
  4. Given OA=r=5 cmOA = r = 5\text{ cm} and AM=4 cmAM = 4\text{ cm}: 52=OM2+425^2 = OM^2 + 4^2 25=OM2+1625 = OM^2 + 16 OM2=9OM^2 = 9 OM=3 cmOM = 3\text{ cm}.
  5. The distance of the chord from the center is 3 cm3\text{ cm}.

Explanation:

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

Problem 4:

Two chords ABAB and CDCD of a circle are parallel and a diameter is perpendicular to them. If AB=6 cmAB = 6\text{ cm}, CD=8 cmCD = 8\text{ cm} and the radius is 5 cm5\text{ cm}, find the distance between the chords if they lie on the same side of the center.

Circle with center O and two parallel chords AB and CD on the same side of the center.

Solution:

  1. Let OO be the center. Let OM⊥ABOM \perp AB and ON⊥CDON \perp CD. MM and NN are midpoints.
  2. AM=62=3 cmAM = \frac{6}{2} = 3\text{ cm} and CN=82=4 cmCN = \frac{8}{2} = 4\text{ cm}.
  3. In △OMA\triangle OMA: OM=52−32=25−9=16=4 cmOM = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}.
  4. In △ONC\triangle ONC: ON=52−42=25−16=9=3 cmON = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}.
  5. Distance between chords =OM−ON=4−3=1 cm= OM - ON = 4 - 3 = 1\text{ cm}.

Explanation:

The distance between two parallel chords on the same side of the center is the difference between their respective distances from the center.