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Circles - Equal Chords and Their Distances from the Centre

Grade 9CBSE

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

🔑Concepts

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The distance of a line from a point is the length of the perpendicular segment drawn from the point to the line. In a circle, the distance of a chord ABAB from the center OO is the length of the perpendicular OMOM where MM lies on ABAB. Visually, this is the shortest path from the center to any point on the chord.

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Theorem 1: Equal chords of a circle (or of congruent circles) are equidistant from the center. If two chords ABAB and CDCD have the same length (AB=CDAB = CD), their perpendicular distances from the center OO (say OMOM and ONON) must be equal (OM=ONOM = ON). Visually, this means identical chords 'sit' at the same depth within the circle.

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Theorem 2 (Converse): Chords that are equidistant from the center of a circle are equal in length. If the perpendicular distances from the center OO to two chords ABAB and CDCD are equal (OM=ONOM = ON), then the chords themselves must be equal (AB=CDAB = CD).

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Perpendicular Bisector Property: The perpendicular drawn from the center of a circle to a chord bisects the chord. If OM⊥ABOM \perp AB, then MM is the midpoint of ABAB, meaning AM=MB=12ABAM = MB = \frac{1}{2}AB. This property is fundamental to calculating distances using right-angled triangles.

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Geometric Relationship: In a circle with center OO and chord ABAB, a right-angled triangle ΔOMA\Delta OMA is formed where OAOA is the radius (rr), OMOM is the distance from the center (dd), and AMAM is half the length of the chord (l/2l/2). You can visualize this as a triangle connecting the center, the midpoint of the chord, and one endpoint of the chord.

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Chord Length vs. Distance: For a given circle, chords that are longer are closer to the center. The diameter is the longest chord of a circle and its distance from the center is 00. Conversely, smaller chords are further away from the center.

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Congruent Circles Application: All theorems regarding equal chords and their distances apply across congruent circles (circles with equal radii). If two separate circles have the same radius, equal chords in both circles will be at the same perpendicular distance from their respective centers.

📐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.