Perpendicular from the Centre to a Chord: Definitions, Theorems, Examples

In this article, we will learn about what is Perpendicular from the Centre to a Chord. A circle is a closed figure with a round shape whose boundary points are all equidistant from the centre. The radius of a circle is the distance between any two locations on its circumference and the centre. Wheels, bangles, clock dials, keychains, shirt buttons and other round things are common in our daily life.

The chord of a circle is the line segment joining any two points on its boundary or circumference. The diameter, which is double the radius, is the longest chord in a circle. Let us learn a property related to the perpendicular drawn from the centre to a chord. Continue reading to know more.

Chord of a Circle

A circle refers to a set of points in a plane located at the same distance from a fixed point. The following terms are used to describe the circle:

Radius: The fixed distance between the circle’s centre and any point on the boundary or circumference of the circle.
Chord: The line segment that connects any two points on the circumference of a circle. Chords can be found in an infinite number of combinations in circles.
Diameter: The diameter is double the radius, and it is the circle’s longest chord, passing through its centre. There is an infinite number of diameters in a circle.

In the above figure, \(O\) is the centre of the circle. \(DC,\, EF\) and \(AB\) are the chords of the circle. \(AB\) is the longest chord called the diameter. \(OR\) is the radius of the circle, which is half the length of \(AB\). \(OP\) is the perpendicular dropped from the centre of the circle to the chord \(EF\).

If the chords \(DC\) & \(EF\) are of equal length, they will be at an equal distance from the circle’s centre.
Equal chords are those that are equidistant from the centre of a circle. On the other hand, unequal chords will not be equidistant from the centre of the circle.

Properties of Chords of a Circle

1. A circle chord divides the circle into two regions, the major and minor segments, respectively.

In the figure, \(AB\) is a chord. \(ACB\) is the minor segment, and \(ADB\) is the major segment. When \(AB\) becomes the diameter, the segments will be equal, and each segment will be a semicircle.

2. When both sides of a chord are extended indefinitely, it becomes secant.

3. Arc length is the distance between two points along the circumference of the circle. So, we can say that the chord length belonging to the greater arc length is greater.

Perpendicular From the 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 from the centre of a circle to a chord bisects the chord.
Proof: Consider a circle with the centre \(O\) and a chord \(AB\) is drawn on the circle.

Construction: Draw the perpendicular \(OM\) to the chord \(AB\) Join \(OA\) and \(OB.\)
In \(\Delta OAM\) and \(\Delta OBM\),
\(OA = OB\) (radius of the circle)
\(OM = OM\) (Common side)
\(\angle OMA = \angle OMB\) (Each angle being equal to \(90^\circ\) as \(OM\) is perpendicular to \(AB\))
Therefore, \(\Delta OAM \cong \Delta OBM\) (By \(SAS\) congruency rule)
\(AM = MB\) (C.P.C.T.)
Thus, the perpendicular \(OM\) bisects the chord \(AB.\)

Theorem: Prove that a line drawn through the centre of a circle that bisects the chord is perpendicular to the chord.
Proof: Consider \(AB\) be the chord of a circle with the centre \(O.\)

Construction: Join \(O,\,M\) and Join \(O,\,A\) and \(O,\,B.\)
In \(\Delta OAM\) and \(\Delta OBM\),
\(OA = OB\) (radius of the circle)
\(AM = BM\) (Given)
\(OM = OM\) (Common side)
Therefore, \(\Delta OAM \cong \Delta OBM\) (By \(SSS\) congruency rule)
Now, \(\angle OMA = \angle OMB\) (CPCT)
Now, \(AB\) is a straight line segment.
\(\angle OMA + \angle OMB = 180^\circ \)
Thus, \(2\angle OMA = 180^\circ \Rightarrow \angle OMA = 90^\circ \)
Or, \(2\angle OMB = 180^\circ \Rightarrow \angle OMB = 90^\circ \)
Hence, the angles \(\angle OMA\) and \(\angle OMB\) are right angles.
Thus, the line drawn from the centre is perpendicular to the chord.

Theorem: If a circle’s diameter bisects two chords of the circle, then those two chords are parallel to each other.

Let us consider \(AB\) and \(CD\) to be the two chords of the circle.
The diameter \(PQ\) passes through the circle with the centre of \(O\).
The diameter \(PQ\) bisects the two chords \(AB,\,CD\) at points \(R\) and \(S\). 
We know that when the line passing through the circle’s centre bisects the line, the line is perpendicular to the chord.

Thus, \(\angle ARS = \angle DSR = 90^\circ ,\;\angle BRS = \angle DSR = 90^\circ \)
Therefore, \(AB \parallel CD\) and cut by the transversal \(PS\) as the interior angles on the same side of the transversal and the alternate angles are equal. 

Solved Examples – Perpendicular From the Centre to a Chord

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

A perpendicular \(OM\) is drawn, as shown in the figure.
In the smaller circle with the centre \(O,\,OM\) is perpendicular to \(BC.\)
We know that a perpendicular divides the chord into two equal parts if drawn from the centre.
Thus, \(BM = CM\) —–(i)
Likewise, for the larger circle, \(OM ⊥ AD.\)
We know that a perpendicular divides the chord into two equal parts if drawn from the centre.
\(AM = DM\) ——(ii)
Subtract (i) from (ii),
\(AM – BM = DM – CM\)
\(AB = CD\)
Hence, proved.

Q.2. Find the length of \(AB\) and \(DC\) when \(OE\) is \(5\;\rm{cm}\), \(AB = DC\), and the radius of the circle is \(13\;\rm{cm}.\)

Ans: Given, \(AO = 13\;\rm{cm},\;OE = 5\;\rm{cm}\)
Let us consider \(△AOE.\)
It is a right-angled triangle as \(OE\) is perpendicular to \(AB.\)
So, using the Pythagoras theorem we have,
\(AO^2 = AE^2 + OE^2\)
Thus,
\(13^2 = AE^2 + 5^2\)
\(⇒ AE^2 = 13^2 -5^2\)
\( \Rightarrow AE = \sqrt {169 – 25} = 12\)
Thus the length of \(AE = 12\;\rm{cm}\)
We know that a perpendicular divides the chord into two equal parts if drawn from the centre.
Thus, \(AB = DC = 2 × 12\,\rm{cm} = 24\,\rm{cm}.\)

Q.3. Two chords are at equal distance from the centre, and the length of one chord is \(8\) units, then find the length of the other chord.
Ans: Given that two chords are at an equal distance from the centre of a circle.
We know that two chords that are at the same distance from the centre are equal in length.
Given the length of one chord is \(8\) units.
Hence, the length of the other chord is \(8\) units.

Q.4. Observe the given figure (\(OM\) is perpendicular on \(AB\)); find the length of the chord \(AB.\)

Ans: In the figure, the line \(OM \bot AB.\)
We know that a perpendicular divides the chord into two equal parts if drawn from the centre.
\( \Rightarrow AM = BM = \frac{{AB}}{2}\)
\( \Rightarrow \frac{{AB}}{2} = 5\)
\( \Rightarrow AB = 10\) units.
Hence, the length of the chord \(AB = 10\) units.

Q.5. Find the length of \(AB\) from the given figure.

Ans: Given, \(CO = 30\,\rm{cm},\;OA = 34\,\rm{cm}\)
Let us consider \(△AOC.\)
It is a right-angled triangle as \(OC\) is perpendicular to \(AB.\)
So, using the Pythagoras theorem we have,
\(AO^2 = AC^2 + OC^2\)
Thus,
\(34^2 = AC^2 + 30^2\)
\(⇒ AC^2 = 34^2 -30^2\)
\( \Rightarrow AC = \sqrt {1156 – 900} = \sqrt {256} = 16\)
We know that a perpendicular divides the chord into two equal parts if drawn to the chord from the centre.
\(⇒ AB = AC + CB = 2AC = 2 × 16 = 32\,\rm{cm}\)
Hence, the length of \(AB\) is \(32\;\rm{cm}.\)

Summary

In this article, we have learnt the definition of a circle and its parts, such as radius, diameter, and chord. We learnt that diameter is the longest chord in a circle, and it is double the radius. We have learnt the theorems related to the chord of a circle when perpendiculars are erected from the centre of the circle to the chord. In the end, we solved some examples related to this theorem.

Frequently Asked Questions

We have provided some frequently asked questions here:

Q.1. What is a line segment that passes through the centre of a circle?
Ans: A diameter is a line segment that passes through the centre of a circle. A circle’s diameter is double its radius’s length.

Q.2. What is the distance around a circle called?
Ans: The distance around a circle is called the circumference. The circle’s circumference or perimeter is the measurement of the circle’s boundary.

Q.3. What are the properties of chords?
Ans: The following are the properties of arcs and chords:
1. The straight line is drawn from the centre of a circle to bisect a chord is perpendicular to the chord.
2. The perpendicular to a chord from the centre of the circle bisects the chord.
3. Equal chords of a circle are equidistant from the chord.
4. Chords of a circle that are equidistant from the centre of the circle are equal.

Q.4. What is the chord and perpendicular theorem?
Ans: A chord is a line segment joining any two points on the circumference of a circle. The chord and perpendicular theorem state that the perpendicular drawn from the centre of a circle bisects the chord.

Q.5. What is the relationship between a chord and an arc?
Ans: A chord divides a circle into minor and major segments. The segments are enclosed by the chord on one side and the arc on the other side. If two chords are equal in measure, their corresponding minor arcs are also equal.

We hope this detailed article on the perpendicular from centre to the chord helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!