krit.club logo

I'm Up and Down, and Round and Round - Distance of Chords from the Centre

Grade 9CBSE

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

🔑Concepts

•

The distance of a line from a point is the length of the perpendicular drawn from the point to the line. In a circle, this point is usually the centre OO.

•

Theorem: The perpendicular from the centre of a circle to a chord bisects the chord. If OM⊥ABOM \perp AB, then AM=MB=12ABAM = MB = \frac{1}{2}AB.

•

Theorem: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

•

Theorem: Equal chords of a circle (or of congruent circles) are equidistant from the centre. If AB=CDAB = CD, then their perpendicular distances from the centre are equal.

•

Theorem: Chords equidistant from the centre of a circle are equal in length.

•

Pythagoras Theorem application: In a right-angled triangle formed by the radius (rr), the distance from the centre (dd), and half the chord length (l2\frac{l}{2}), the relation is r2=d2+(l2)2r^2 = d^2 + (\frac{l}{2})^2.

📐Formulae

AM=MB=12ABAM = MB = \frac{1}{2}AB

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

d=r2−(l2)2d = \sqrt{r^2 - \left(\frac{l}{2}\right)^2}

l=2r2−d2l = 2\sqrt{r^2 - d^2}

💡Examples

Problem 1:

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

Solution:

Given: Chord length l=16 cml = 16\text{ cm}, Distance from centre d=6 cmd = 6\text{ cm}. Let the chord be ABAB and the perpendicular from centre OO be OMOM. AM=12×AB=12×16=8 cmAM = \frac{1}{2} \times AB = \frac{1}{2} \times 16 = 8\text{ cm} In right ΔOMA\Delta OMA, using Pythagoras theorem: OA2=OM2+AM2OA^2 = OM^2 + AM^2 r2=62+82r^2 = 6^2 + 8^2 r2=36+64r^2 = 36 + 64 r2=100r^2 = 100 r=100=10 cmr = \sqrt{100} = 10\text{ cm} Radius of the circle is 10 cm10\text{ cm}.

Explanation:

We first find the half-length of the chord because the perpendicular from the centre bisects it. Then, we apply the Pythagoras theorem to the triangle formed by the radius, the distance, and the half-chord.

Problem 2:

Find the length of a chord which is at a distance of 5 cm5\text{ cm} from the centre of a circle of radius 13 cm13\text{ cm}.

Solution:

Given: Radius r=13 cmr = 13\text{ cm}, Distance d=5 cmd = 5\text{ cm}. Let the chord length be ll. We know: (l2)2=r2−d2\left(\frac{l}{2}\right)^2 = r^2 - d^2 (l2)2=132−52\left(\frac{l}{2}\right)^2 = 13^2 - 5^2 (l2)2=169−25\left(\frac{l}{2}\right)^2 = 169 - 25 (l2)2=144\left(\frac{l}{2}\right)^2 = 144 l2=144=12 cm\frac{l}{2} = \sqrt{144} = 12\text{ cm} l=12×2=24 cml = 12 \times 2 = 24\text{ cm} The length of the chord is 24 cm24\text{ cm}.

Explanation:

Using the relation between radius, distance, and chord length, we first calculate the half-chord length and then double it to find the full length of the chord.

Problem 3:

Two equal chords ABAB and CDCD of a circle with centre OO are such that AB=10 cmAB = 10\text{ cm}. If the distance of ABAB from OO is 4 cm4\text{ cm}, what is the distance of CDCD from OO?

Solution:

Given: AB=CD=10 cmAB = CD = 10\text{ cm}. Distance of ABAB from O=4 cmO = 4\text{ cm}. According to the theorem: 'Equal chords of a circle are equidistant from the centre.' Since AB=CDAB = CD, the distance of CDCD from the centre OO must be equal to the distance of ABAB from OO. Therefore, distance of CD=4 cmCD = 4\text{ cm}.

Explanation:

This problem uses the property that equal chords are always at the same distance from the centre of the circle.