# Important Questions for Chapter 10 - Circles

The free set of important questions carries step-by-step explanation and solution. Our math subject experts with years of experience have studied the CBSE exam patterns very carefully and come up with these set of questions based on the current CBSE syllabus.

**Important questions based on NCERT syllabus for Chapter 10 - Circles:**

*Question 1*: Prove that the right bisector of a chord of a circle, bisects the corresponding arc of the circle.

*Solution*:

Let AB be a chord of a circle having its centre at O.

Let PQ be the right bisector of the chord AB, intersecting AB at L and the circle at Q.

Since the right bisector of a chord always passes through the centre, so PQ must pass through the centre O.

Join OA and OB.

OA = OB [Each equal to the radius]

∠ALO = ∠BLO [Each equal to 90∘]

OL=OL [Common]

∴ΔOAL ≅ ΔOBL [By RHS congruency criterion]

⇒ ∠AOL = ∠BOL [C.P.C.T]

∴AQ = BQ [Arcs subtending equal angles at the centre are equal]

*Question 2*: In the adjoining figure, AB and AC are two equal chords of a circle with center O. Show that O lies on the bisector of ∠BAC.

*Solution*:

AB and AC are equal chords of a circle with center O and O has been joined with A.

To prove : ∠BAO=∠CAO.

*Construction* : Join OB and OC.

*Proof*:

In ΔOAB and ΔOAC, we have

(i) AB = AC (Given)

(ii) OB = OC (Radii of the same circle)

(iii) OA = OA (Common side)

∴ ΔOAB ≅ ΔOAC (By SSS congruency criterion)

Hence, ∠BAO = ∠CAO (By C.P.C.T)

Hence the center of the circle lies on the bisector of ∠BAC.

*Question 3*: Prove that if a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic.

*Solution*:

Consider a quadrilateral ABPQ, such that ∠ABP+∠AQP=180∘ and ∠QAB+∠QPB=180∘

To prove: The points A, B, P and Q lie on the circumference of a circle.

Assume that point P does not lie on a circle drawn through points A, B and Q.

Let the circle cut QP at point R. Join BR.

∠QAB + ∠QRB = 180∘ [opposite angles of cyclic quadrilateral.]

∠QAB + ∠QPB = 180∘ [given]

∴ ∠QRB = ∠QPB

But this cannot be true since ∠QRB = ∠QPB + ∠RBP (exterior angle of the triangle)

∴ Our assumption that the circle does not pass through P is incorrect and A, B, P and Q lie on the circumference of a circle.

∴ ABPQ is a cyclic quadrilateral.