Equal Chords and their Distances from the Centre: We may have observed many round objects in daily life, such as dials of many clocks, coins, keyrings, wheels of a vehicle, bangles, buttons of shirts, etc. A circle is a closed object, which is round.

The chord of the circle is the line segment joining between any two points on the circumference. We have different theorems associated with the chords of the circle. We shall discuss some of these theorems in this article.

Chord of a Circle

The set of collections of all the points in a plane at a fixed distance from a fixed point in the plane is called a circle. The various terms related to the circle are discussed below:

• Radius: The fixed distance between the centre and any point on the circle.
• Diameter: It is the chord of the circle, passing through the centre, and it is the longest chord.
• Tangent: The line touching the circle at only one point.
• Arc: The part of the boundary of a circle.
• Chord: The line segment joining any two points on the circle.

The chord of the circle is the line segment joining between two points on the circumference. A circle can have an infinite number of chords.

In the above figure, the line segment \(QR\) is the chord of a circle, as it is joining the two points \(Q\) and \(R\) on the circle.

Equal Chords and Their Distances from Centre

Let \(P\) be any point and let \(AB\) be any line, as shown in the figure. There are infinite numbers of points on a line. We will have infinitely many line segments \(P{L_1},P{L_2},PM,P{L_3},P{L_4},\) etc., by joining these points to \(P.\)

From all of these line segments, the perpendicular \((PM)\) from \(P\) to \(AB\) is the least in length. The line segment having the least length \((PM)\) gives the distance of the point \(P\) from the line \(AB.\)

The perpendicular to a line from the point gives the distance of the line from the point.

As we know that a circle can have an infinite number of chords. As discussed above, we can say that the chord which is nearer to the centre of a circle is the longest chord. On the contrary, the chord that is away from the centre of a circle is the least. That’s why we can say that diameter is the longest chord of a circle as it passes through the centre of a circle.

Perpendicular from Centre to the Chord

We know that the shortest line drawn from the centre of a circle to the chord is perpendicular to the chord. The perpendicular line drawn to the chord from the centre of a circle bisects the chord.

Theorem: The perpendicular bisects the chord if it is drawn on the chord from the centre.

Proof: Take a circle with the centre \(“O”,\) and \(AB\) be the chord of a circle.

And draw the perpendicular \(OM\) to the chord \(AB.\)
Join \(OA\) and \(OB.\)

In triangles \(OAM\) and \(OBM,\)

\(OA=OB\) (radius of the same circle)

\(OM=OM\) (Common side)

\(∠OMA=∠OMB\) (Each angle being equal to \({\rm{9}}{{\rm{0}}^{\rm{o}}}\))

By SAS congruency rule, \(∆OAM≅∆OBM.\)

Corresponding parts of the congruent triangles (C.P.C.T)

\(AM=MB\)

Thus, the perpendicular \(OM\) bisects the chord \(AB.\)

Equal Chords Subtend Equal Angles at the Centre

Proof: Consider a circle with centre \(O.\)

Let us assume the chords of the circles \(AB\) and \(CD\) are equal.

\(AB = CD……..{\rm{(i)}}\)

In triangles, \(AOB\) and \(COD,\)

\(OA=OC\) (Radius of the same circle)

\(OB=OD\) (Radius of the same circle)

\(AB=CD\) (Given equal chords)

By the SSS congruency rule,

\(∆AOB≅∆COD\)

Corresponding parts of congruent triangles (C.P.C.T)

\(∠AOB=∠COD\)

Therefore, equal chords make the same angles at the centre.

Equal Chords and Their Distances from the Centre: Theorem & Proof

Theorem: Equal chords of circles with the same radius are at an equal distance from the centre.

Proof: Consider a circle with the centre \(“O”.\)
We have to prove \(OL=OM.\)

Here, the chords \(PQ\) and \(RS\) drawn in the circle are equal in length.

Let us draw perpendiculars \(OL\) and \(OM\) two the chords \(PQ\) and \(RS,\) respectively.

We know that a perpendicular bisects the chord if it is drawn to the chord from the centre.

\(PQ=RS\)

\( \Rightarrow \frac{1}{2}PQ = \frac{1}{2}RS\)

\(PL=RM\)

In triangles \(OPL\) and \(ORM,\)

\(OP=OR\) (radius of the same circle)

\(PL=RM\) (Proved above)

\(∠OLP=∠OMR\) (Both are equals to \({\rm{9}}{{\rm{0}}^{\rm{o}}}\))

By using the SAS congruency rule.

\(∆OLP≅∆OMR\)

The corresponding parts of congruent triangles (C.P.C.T)

\(OL=OM\)

Hence, we can say that two equal chords are at equal distances from the centre of a circle.

Converse Theorem

Two chords that lie at an equal distance from the centre of a circle are equal in length.

Proof: Take a circle with centre \(O.\)

We have to prove \(PQ=RS.\)

Let us draw perpendiculars \(OL\) and \(OM\) on the two chords \(PQ\) and \(RS,\) respectively.

Here, the perpendicular distances of the chords \(PQ\) and \(RS\) from the centres drawn in the circle \(OL\) and \(OM\) are equal.

In triangles \(OPL\) and \(ORM,\)

\(OP=OR\) (radius of the same circle)

\(OL=OM\) (Given that perpendicular distances are equal)

\(∠OLP=∠OMR\) (Both are equals to \({\rm{9}}{{\rm{0}}^{\rm{o}}}\))

By using the SAS congruency rule.

\(∆OLP≅∆OMR\)

The corresponding parts of congruent triangles (C.P.C.T)

\(PL = RM……..\left( {\rm{i}} \right)\)

We know that a perpendicular bisects the chord if it is drawn to the chord from the centre.

Thus, \(OL\) bisects the chord \(PQ,\) and \(OM\) bisects the chord \(RS.\)

\(PL = \frac{{PQ}}{2} \ldots {\rm{(ii)}}\) and \(RM = \frac{{RS}}{2} \ldots {\rm{(iii)}}\)

From \({\rm{(i)}},{\rm{(ii)}}\) and \({\rm{(iii),}}\)

\(\frac{{PQ}}{2} = \frac{{RS}}{2}\)

\(PQ = RS\)

Hence, we can say that two chords at equal distances are equal in length.

Solved Examples on Equal Chords and their Distances from Centre

Q.1. A line drawn to concentric circles intersects two concentric circles at \(A, B, C\) and \(D,\) prove that \(AB=CD.\)
Ans: Given a line intersects at points \(A, B, C, D\) of the concentric circles.

Draw a perpendicular \(OM\) as shown in the figure.
For the smaller circle of centre, \(“O”, OM⊥BC.\)
We know that a perpendicular bisects the chord if it is drawn to the chord from the centre.
Thus, \(BM = CM……..\left( {\rm{i}} \right)\)
Similarly, for the larger circle, \(OM⊥AB.\)
We know that a perpendicular bisects the chord if it is drawn to the chord from the centre.
\(AM = DM……..\left( {\rm{ii}} \right)\)
Subtract \(\left( {\rm{i}} \right)\) from \(\left( {\rm{ii}} \right)\)
\(AM – BM = DM – CM\)
\(AB = CD\)
Hence, proved.

Q.2. Prove that a line drawn through the centre of a circle that bisects the chord is perpendicular to the chord.
Ans: Consider a circle with the centre \(“O”\) and \(AB\) be the chord of a circle.

And draw the perpendicular \(OM\) to the chord \(AB.\)
Join \(OA\) and \(OB.\)
In triangles \(OAM\) and \(OBM,\)
\(OA=OB\) (radius of the same circle)
\(OM=OM\) (Common side)
\(AM=BM\) (Given)
By S.S.S congruency rule, \(∆OAM≅∆OBM.\)
Corresponding parts of the congruent triangles (C.P.C.T)
\(∠OMA=∠OMB\)
Now, \(AB\) is a straight line. Hence, each of the angles \(∠OMA\) and \(∠OMB\) are right angles.
Thus, the line drawn from the centre is perpendicular to the chord.

Q.3. Find the distance of the chord from the centre, such that two chords are equal in length, and the distance of the other chord from the centre is \({\rm{5}}\,{\rm{cm}}{\rm{.}}\)
Ans: We know that two equal chords are equidistant from the centre.
That is, they have the same perpendicular distances.
Given, one chord is at the distance of \({\rm{5}}\,{\rm{cm}}{\rm{.}}\)
So, the other chord is also at a distance of \({\rm{5}}\,{\rm{cm}}\) from the centre.

Q.4. Two chords are at equal distance from the centre, and the length of one chord is \(6\) units, then find the length of the other chord.
Ans: Given two chords are at equal distance from the centre of a circle.
We know that two chords that are equidistant from the centre are equal in length.
Given the length of one chord is \(6\) units.
So, the length of the other chord is \(6\) units.

Q.5. Observe the given figure (\(OM\) is perpendicular on \(AB\)); find the length of the chord \(AB.\)

Ans: In the figure, the line \(OM⊥AB.\)
We know that a perpendicular bisects the chord if it is drawn to the chord from the centre.
\( \Rightarrow AM = BM = \frac{{AB}}{2}\)
\( \Rightarrow \frac{{AB}}{2} = 5\)
\( \Rightarrow AB = 10\) units.
Hence, the length of the chord \( \Rightarrow AB = 10\) units.

Summary

In this article, we have discussed the definitions of a circle and the parts of the circle such as radius, diameter, arc and chord of the circle. 

This article gives the theorem regarding the equal chords and their distances from the centre, such as equal chords are equidistant from the centre. Conversely, two chords that are equidistant from the centre are equal etc. This article gives the solved examples which help us to understand the concepts easily.

FAQs

Q.1. What will be the distance of two equal chords from the centre?
Ans: The distance of two equal chords from the centre of the circle is equal.

Q.2. What will be the length of chords if they are at the same distance from the centre of a circle?
Ans: The two chords are equal in length if they are equidistant from the centre of a circle.

Q.3. Are equal chords of a circle make equal angles at the centre?
Ans: Equal chords of a circle make equal angles at the centre of a circle.

Q.4. What is “equal chords and distance from the centre theorem”?
Ans: The theorem states that equal chords of a circle are at the same distance from the centre of a circle.

Q.5. What is the perpendicular and chord theorem?
Ans: The perpendicular drawn to the chord from the centre of a circle bisects the chord.

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