 # Circles Class 9

Circles class 9 – The collection of all points equidistant from a fixed point in a plane is called a circle. Let us discuss in detail about the definitions and terminologies used in a concept of the circle.

## Circles Class 9 Notes

• A plane is divided by the circle into 3 parts, namely, the interior and exterior of the circle and the circle itself.
• A chord is a line passing from one point to another on the circumference of a circle.
• Diameter is a chord that passes through the centre.
• The perimeter of a circle is known as the circumference.
• Chords equal in length subtend equal angles at centre. Similarly, the chords are equal if the angles subtended by chords are equal at the centre.
• The perpendicular drawn on a chord from centre bisects the chord. Similarly, the line is perpendicular if it is drawn from the centre to bisects the chord.
• Only one circle can pass through 3 non-collinear points.
• Chords equal in length are equidistant from the centre. Similarly, Chords are equal if they are equidistant from the centre.
• In a cyclic quadrilateral, the sum of pair of angles opposite to each other is 180 degree.
• Angles subtended by the same line segment are equal.
• If chords of a circle are equal in length, then the corresponding arcs will be congruent. Similarly, if arcs are congruent, then the corresponding chords will be equal.

## Circles Class 9 Examples

Example 1:

An arc of a circle is given. Complete the circle.

Solution: Let assume that PQ is the given arc of a circle. Now, we have to complete the circle for the given arc. It means that we need to find the radius and the centre point of a circle.

Now, take the point R on the circle, and join the points PR and RQ. Now, by using the theorem: “There is one and only one circle passing through the three given non-collinear points”. Now, construct a circle based on this theorem to find the centre and radius. After obtaining the radius and centre, complete the circle for the given arc.

Example 2:

Prove that the chords are equal if two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection.

Solution: Given: AB and CD are the two chords of a circle with centre “O”. The intersection of these two chords is “E”. Now, PQ is the diameter of a circle, which passes through the point, such that  ∠AEQ = ∠DEQ.

Now, we need to prove AB = CD. To prove this, draw the perpendiculars OL and OM on the chords AB and CD respectively.

By using the angle sum property of a triangle, we can write:

∠LOE = 180° – 90° – ∠LEO = 90° – ∠LEO

Therefore,

=90° – ∠AEQ = 90° – ∠DEQ

= 90° – ∠MEO = ∠MOE

From the triangles, OLE and OME,

∠LEO = ∠MEO

∠LOE = ∠MOE

EO = EO (Common)

Hence, ∆ OLE ≅ ∆ OME, Which means that OL = OM

Thus, the chord AB and CD are equal.

Hence, proved.

Example 3:

Two circles intersect at two points A and B, and AC and AD are the diameters of the circle. Show that the point B lies on the line segment DC. Solution:

From the given figure, join the points A and B.

Now, ∠ABD = 90° and ∠ABC = 90° (Angle in a semicircle)

Therefore, ∠ABD + ∠ABC = 90° + 90° = 180°

Now, DBC is a line. It means that point B lies in the line segment DC.

Hence, proved.

## Video Lessons on Circles

### Introduction to Circles ### Parts of a Circle ### Area of a Circle  ### Practice Problems

1.  Prove that the segments of one chord are equal to corresponding segments of the other chord if two equal chords of a circle intersect within the circle.
2. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, then determine ∠ BCD. Also, if AB = BC, find ∠ECD.
3. Show that a cyclic parallelogram is a rectangle.

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