**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:Â **

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.Â

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:Â **

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 thatÂ âˆ AEQ = âˆ DEQ.Â

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:

âˆ LOE = 180Â° â€“ 90Â° â€“ âˆ LEO = 90Â° â€“ âˆ LEO

Therefore,

=90Â° â€“ âˆ AEQ = 90Â° â€“ âˆ DEQ

= 90Â° â€“ âˆ MEO = âˆ MOE

From the triangles, OLE and OME,

âˆ LEO = âˆ MEO

âˆ LOE = âˆ MOE

EO = EO (Common)

Hence, âˆ† OLE â‰… âˆ† OME, Which means that OL = OM

Thus, the chord AB and CD are equal.Â

Hence, proved.

**Example 3:Â **

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, âˆ ABD = 90Â° and âˆ ABC = 90Â° (Angle in a semicircle)

Therefore, âˆ ABD + âˆ ABC = 90Â° + 90Â° = 180Â°

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

Hence, proved.

### Practice Problems

- Â 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.
- ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If âˆ DBC = 70Â°, âˆ BAC is 30Â°, then determine âˆ BCD. Also, if AB = BC, find âˆ ECD.
- Show that a cyclic parallelogram is a rectangle.Â

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