Class 9 Maths Chapter 10 Practice Question Answers - Circles

Q1. In the figure, OD is perpendicular to chord AB of a circle whose centre is O. If BC is a diameter, prove that CA = 2OD.

Solution: Given: BC is a diameter of a circle with centre O and OD⊥AB.

To prove: AC parallel to  OD and AC = 2 × OD

Construction: Join AC.

Proof: We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Here, OD⊥AB

D is the mid point of AB.

i.e., AD=BD

Also, O is the mid point of BC

.i.e., OC=OB

Now, in ΔABC, we have:

D is the mid point of AB and O is the mid point of BC.

According to the mid point theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.

i.e., OD∥AC and OD=1/2AC

⇒ AC=2×OD

Hence, proved.

 Q2. l is a line intersecting two concentric circles having common centre O, at A, B, C and D. Prove that AB = CD.
 Solution: 

Given: A line intersects two concentric circles (circles with the same center) with center O at A, B, C, and D

To prove: AB = CD

Construction: Draw OM ⊥ BC.

Proof: The perpendicular drawn from the centre of a circle to a chord bisects the chord.

AM = DM ---(1)

BM = CM -----(2)

Subtracting (2) from (1), we get

AM - BM = DM - CM

⇒ AB = CD

Hence Proved

Q3. AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E. Prove that EB = ED.

Solution: From O draw OP⊥AB and OQ⊥CD. Join OE.

Given, AB = CD.

Since equal chords of a circle are equidistant from the centre, OP = OQ.

Now, in right triangles OPE and OQE,

OE = OE (common)

OP = OQ (proved above)

⟹△OPE≅△OQE (R.H.S.)

∴ PE = QE (C.P.C.T.)

⟹ PE−1/2AB=QE−1/2CD(because AB=CD (given))

⟹PE−PB=QE −QD

⟹EB=ED

⟹BE+AB=ED+CD (∵ AB=CD)

⟹AE =CE

Hence the given statement is true.

Q4. If O be the centre of the circle, find the value of ‘x’ in each of the following figures.

Solution: 

(i) ∵ OA = OB                 [Radii of the same circle]
∴ ∠A = ∠B                 [Angles opposite to equal side in a triangle are equal]
In ΔABC, ∠A + ∠B + ∠O = 180º
∴ x + x + 70º = 180º                 [∵ ∠O = 70º (given) and ∠A = ∠B]
⇒ 2x + 70º = 180º
⇒ 2x = 180º - 70º = 110º
⇒ x= (110⁰/2)= 55º
Thus, x = 55°

(ii) In ΔAOC, ∠A + ∠ACO + ∠AOC = 180º
⇒ 40º + ∠ACO + 90º = 180º
⇒ ∠ACO = 180º - 40º - 90º = 50º
∵ AB is a diameter.
∴ ∠ACB = 90º                 [Angle in a semicircle]
∴ 50º + x = 90º
⇒ xº = 90º - 50º = 40º
Thus, x = 40º

(iii) ∵ ∠AOC + ∠COB = 180º                 [Linear pairs]
∴ 120º + ∠COB = 180º
⇒ ∠COB = 180º - 120º = 60º
∵ The arc CB is subtending ∠COB at the centre and ∠CDB at the remaining part.
∴ ∠CDB = (1/2)∠COB
⇒ x= (1/2)(60º) = 30º
⇒ x = 30º

(iv) In ΔAOC,
∵ AO = OC                 [Radii of the same circle]
∴ ∠OAC = ∠OCA                [Angles opposite to equal sides are equal]
⇒ ∠OAC = 50º
∴ Exterior ∠COB = 50º + 50º = 100º.
Now, the arc BC is subtending ∠BOC at the centre and ∠BDC at the remaining part of the circle.
∴ ∠BDC = (1/2)∠BOC
⇒ x= (1/2)(100º) = 50º
Thus, x = 50º

(v) In ΔOAC, OA = OC                 [Radii of the same circle]
∴ ∠AOC = ∠ACO                [∵ Angles opposite to equal sides are equal]
Now, ∠AOC + ∠ACO + ∠OAC = 180º
⇒ ∠AOC + ∠ACO + 50º = 180º
⇒ ∠AOC + ∠ACO = 180º - 50º = 130º
⇒ ∠AOC = ∠ACO = (130⁰/2) = 65º
Now, ∠AOB + ∠AOC = 180º                [Linear pairs]
∴ ∠AOB + 65º = 180º
⇒ ∠AOB = 180º - 65º = 125º
∵ The arc AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle.  
∴ ∠ADB = (1/2) ∠AOB
⇒ x= (1/2)(125º) = 62 (1/2)
∴ x= 62(1/2)º

(vi) ∵ ∠BDC = ∠BAC                [Angles in the same segment]
∴ ∠BDC = 40º
Now, in ΔBDC, we have ∠BDC + ∠CBD + ∠BCD = 180º
∴ 40º + 80º + x = 180º
⇒ 120º + x = 180º
⇒ x = 180º - 120º = 60º
Thus, x = 60º

Q5. In the adjoining figure, O is the centre of the circle. Prove that ∠ XOZ = 2(∠ XZY + ∠ YXZ).

 Solution: Let us join OY.
∵ The arc XY subtends ∠XOY at the centre and ∠XZY at a point Z on the remaining part of the circle.

∴ ∠XOY = 2∠XZY                …(1)
Similarly, ∠YOZ = 2∠YXZ                …(2)
Adding (1) and (2), we have ∠XOY + ∠YOZ = 2∠XZY + 2∠YXZ
⇒ ∠XOZ = 2[∠XZY + ∠YXZ]

Q6. Show that the sum of the opposite angles of a cyclic quadrilateral is 180º.
 Solution: We have a cyclic quadrilateral. Let us join AC and BD. Since, angles in the same segment are equal.
∴ ∠ACB = ∠ADB                …(1)
and ∠BAC = ∠BDC                …(2)
Adding (1) and (2), we have

∠ACB + ∠BAC = ∠ADB + ∠BDC
⇒ ∠ACB + ∠BAC = ∠ADC
Adding ∠ABC to both sides, we have ∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABC
But, ∠ACB + ∠BAC + ∠ABC = 180º                 [Sum of the angles of ΔABC = 180º]
∴ ∠ADC + ∠ABC = 180º
⇒ ∠B + ∠D = 180º
Since, ∠A + ∠B + ∠C + ∠D = 360º
⇒ ∠A + ∠C = 360º ∠ 180º = 180º

Q7. Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
 Solution: We have a cyclic quadrilateral ABCD in which the bisectors ∠A, ∠B, ∠C and ∠D for a quadrilateral PQRS.
From ΔABP, we have ∠PAB + ∠PBA + ∠P = 180º                [Sum of the three angles of ΔABP]

⇒ (1/2) ∠A + (1/2)∠B + ∠P = 180º …(1)
From ΔCDR, we have
∠RCD + ∠RDC + ∠R = 180º                [Sum of the three angles of ΔCDR.]
⇒ (1/2)∠C +(1/2)∠D + ∠R = 180°                 …(2)
Adding (1) and (2), we have (1/2)∠A + (1/2)∠B + (1/2)∠C +(1/2)∠D + ∠P + ∠R = 360º
⇒ (1/2) (∠A + ∠B + ∠C + ∠D) + ∠P + ∠R = 360º
⇒ (1/2)(360º) + ∠P + ∠R = 360º                 [∵ ∠A + ∠B + ∠C + ∠D = 360°]
⇒ ∠P + ∠R = 360º -(1/2) (360º) = 180º
Similarly, ∠Q + ∠S = 180º
Thus, the pairs of opposite angles of quadrilateral PQRS are supplementary.
Hence, PQRS is cyclic.

Q8. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD. 

Solution: Consider the chord CD,

As we know, angles in the same segment are equal.

So, ∠CBD = ∠CAD

∴ ∠CAD = 70°

Now, ∠BAD will be equal to the sum of angles BAC and CAD.

So, ∠BAD = ∠BAC + ∠CAD

= 30° + 70°

∴ ∠BAD = 100°

As we know, the opposite angles of a cyclic quadrilateral sum up to 180 degrees.

So,

∠BCD + ∠BAD = 180°

Since, ∠BAD = 100°

So, ∠BCD = 80°

Now consider the ΔABC.

Here, it is given that AB = BC

Also, ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)

∠BCA = 30°

also, ∠BCD = 80°

∠BCA + ∠ACD = 80°

So, ∠ACD = 50° and,

∠ECD = 50°

Q9. In Figure, ∠ABC = 69°, ∠ ACB = 31°, find ∠BDC. 

Solution: As we know, angles in the segment of the circle are equal so,

∠BAC = ∠BDC

Now in the In ΔABC, sum of all the interior angles will be 180°

So, ∠ABC + ∠BAC + ∠ACB = 180°

Now, by putting the values,

∠BAC = 180° – 69° – 31°

So, ∠BAC = 80°

Q10. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Solution: 

Here, it is given that ∠AOB = ∠COD i.e. they are equal angles.

Now, we will have to prove that the line segments AB and CD are equal i.e. AB = CD.

Proof:

In triangles AOB and COD,

∠AOB = ∠COD (as given in the question)

OA = OC and OB = OD ((these are the radii of the circle)

So, by SAS congruency, ΔAOB ≅ ΔCOD.

∴ By the rule of CPCT, AB = CD. (Hence proved).