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Circles - Compare chord lengths using distance from centre and prove related results

Grade 9CBSE

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

🔑Concepts

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Equal chords of a circle (or of congruent circles) are equidistant from the centre. This means if AB=CDAB = CD, then the perpendicular distances OMOM and ONON from the centre OO to these chords are equal (OM=ONOM = ON).

Circle showing two equal chords at equal distances from the center.
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Chords equidistant from the centre of a circle are equal in length. If the perpendicular distances from the centre to two chords are the same, the chords must have the same length (OM=ON  ⟹  AB=CDOM = ON \implies AB = CD).

Circle with two chords at the same distance 'd' from the center.
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Of any two chords of a circle, the longer chord is nearer to the centre. Conversely, the chord which is nearer to the centre is longer.

Circle showing that a longer chord is closer to the center than a shorter chord.
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The perpendicular from the centre of a circle to a chord bisects the chord. This forms a right-angled triangle where the hypotenuse is the radius, allowing for the use of the Pythagoras Theorem: r2=d2+(l/2)2r^2 = d^2 + (l/2)^2.

📐Formulae

Pythagorean relationship: r2=d2+(l2)2r^2 = d^2 + \left(\frac{l}{2}\right)^2

Perpendicular distance from center: d=r2−(l2)2d = \sqrt{r^2 - \left(\frac{l}{2}\right)^2}

Length of the chord: l=2r2−d2l = 2\sqrt{r^2 - d^2}

Radius of the circle: r=d2+(l2)2r = \sqrt{d^2 + \left(\frac{l}{2}\right)^2}

If ABAB and CDCD are chords and OM⊥ABOM \perp AB, ON⊥CDON \perp CD, then: AB=CD  ⟺  OM=ONAB = CD \iff OM = ON

💡Examples

Problem 1:

In a circle of radius 55 cm, two equal chords ABAB and CDCD are drawn. If the length of chord ABAB is 88 cm, calculate the distance of chord CDCD from the center of the circle.

Solution:

Step 1: Identify that since ABAB and CDCD are equal chords (AB=CD=8AB = CD = 8 cm), they are equidistant from the center. Thus, finding the distance of ABAB will give the distance of CDCD. Step 2: Let OO be the center and OM⊥ABOM \perp AB. By the perpendicular bisector theorem, MM bisects ABAB. So, AM=12×8=4AM = \frac{1}{2} \times 8 = 4 cm. Step 3: In right-angled triangle ΔOMA\Delta OMA, we have radius OA=5OA = 5 cm and base AM=4AM = 4 cm. Using Pythagoras Theorem: OA2=OM2+AM2OA^2 = OM^2 + AM^2 Step 4: Substitute the values: 52=OM2+42  ⟹  25=OM2+165^2 = OM^2 + 4^2 \implies 25 = OM^2 + 16 Step 5: Solve for OMOM: OM2=25−16=9  ⟹  OM=9=3 cmOM^2 = 25 - 16 = 9 \implies OM = \sqrt{9} = 3 \text{ cm}. Since equal chords are equidistant, the distance of CDCD from the center is also 33 cm.

Explanation:

This problem uses the property that the perpendicular from the center bisects the chord and applies the Pythagorean theorem to find the distance. The final step relies on the theorem that equal chords are equidistant from the center.

Problem 2:

Two parallel chords ABAB and CDCD of lengths 1010 cm and 2424 cm respectively are on opposite sides of the center of a circle. If the radius of the circle is 1313 cm, find the distance between the two chords.

Solution:

Step 1: Let the center be OO. Draw OM⊥ABOM \perp AB and ON⊥CDON \perp CD. Since AB∥CDAB \parallel CD and they are on opposite sides, the distance between them is MN=OM+ONMN = OM + ON. Step 2: Calculate OMOM for chord AB=10AB = 10 cm. AM=12AB=5AM = \frac{1}{2}AB = 5 cm. In ΔOMA\Delta OMA: OM=OA2−AM2=132−52=169−25=144=12 cmOM = \sqrt{OA^2 - AM^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm} Step 3: Calculate ONON for chord CD=24CD = 24 cm. CN=12CD=12CN = \frac{1}{2}CD = 12 cm. In ΔONC\Delta ONC: ON=OC2−CN2=132−122=169−144=25=5 cmON = \sqrt{OC^2 - CN^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ cm} Step 4: The total distance between the chords is OM+ON=12+5=17OM + ON = 12 + 5 = 17 cm.

Explanation:

The problem requires calculating the individual distances of two different chords from the center using the Pythagorean theorem and then summing those distances because the chords lie on opposite sides of the center.

Problem 3:

In a circle of radius 1010 cm, two chords PQPQ and RSRS are drawn such that they are at distances of 66 cm and 88 cm from the centre respectively. Find the difference in the lengths of the two chords.

Circle with two chords at distances 6 and 8 from the center.

Solution:

  1. For chord PQPQ: radius r=10r = 10 cm and distance d1=6d_1 = 6 cm. Using Pythagoras Theorem: (PQ2)2=r2−d12(\frac{PQ}{2})^2 = r^2 - d_1^2 (PQ2)2=102−62=100−36=64(\frac{PQ}{2})^2 = 10^2 - 6^2 = 100 - 36 = 64 PQ2=64=8\frac{PQ}{2} = \sqrt{64} = 8 cm. So, PQ=2×8=16PQ = 2 \times 8 = 16 cm.

  2. For chord RSRS: radius r=10r = 10 cm and distance d2=8d_2 = 8 cm. (RS2)2=102−82=100−64=36(\frac{RS}{2})^2 = 10^2 - 8^2 = 100 - 64 = 36 RS2=36=6\frac{RS}{2} = \sqrt{36} = 6 cm. So, RS=2×6=12RS = 2 \times 6 = 12 cm.

  3. Difference in lengths = PQ−RS=16−12=4PQ - RS = 16 - 12 = 4 cm.

Explanation:

By calculating the length of each chord using the distance from the centre and the radius, we observe that the chord closer to the centre (66 cm) is longer than the chord further away (88 cm).

Problem 4:

Two equal chords ABAB and CDCD of a circle with centre OO intersect at a point EE within the circle. Prove that the line segment OEOE bisects the angle ∠AEC\angle AEC.

Circle with intersecting chords AB and CD showing perpendiculars from the center.

Solution:

  1. Draw perpendiculars OM⊥ABOM \perp AB and ON⊥CDON \perp CD.
  2. Since AB=CDAB = CD (given), we know OM=ONOM = ON (equal chords are equidistant from the centre).
  3. In △OME\triangle OME and △ONE\triangle ONE:
  • ∠OME=∠ONE=90∘\angle OME = \angle ONE = 90^{\circ} (By construction)
  • OE=OEOE = OE (Common hypotenuse)
  • OM=ONOM = ON (Proved above)
  1. △OME≅△ONE\triangle OME \cong \triangle ONE by RHS congruence rule.
  2. Therefore, ∠OEM=∠OEN\angle OEM = \angle OEN (By CPCT).
  3. Thus, OEOE bisects ∠MEN\angle MEN, which is the same as ∠AEC\angle AEC.

Explanation:

This proof uses the property that equal chords are at equal distances from the centre to establish congruent triangles, leading to the equality of angles.