Exercise 9.1
Multiple Choice Questions (Q. No. 1-5)
Question. 1 If radii of two concentric circles are 4 cm and 5 cm, then length of each chord of one circle which is tangent to the other circle, is
(a) 3 cm             
(b) 6 cm           
(c) 9 cm        
(d) 1 cm
Solution. (b)
Explanation:
Let C₁ and C₂ be the two concentric circles, with centre at O and respective radii being r₁ = 4cm and r₂ = 5cm.
Draw a chord AC of circle C₂, to touch circle C₁ at B.
Join OB.


Solution. (d)
Explanation:
Since, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
i.e.,    ∠AOB + ∠COD = 180°
⟹     ∠COD = 180° – ∠AOB
⟹     ∠COD = = 180° – 125° = 55°
Question. 3 In figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to

Solution. (c)
Explanation:
Here AC is the diameter of the circle.
∴               ∠ABC = 90°                      [ Angle in a semi-circle]
Now, in ΔACB, ∠A + ∠B + ∠C = 180°      [Sum of all interior angles of a triangle is 180°]
⟹     ∠A + 90° + 50° = 180°
⟹             ∠A + 140 = 180
⟹                      ∠A = 180^o – 140° = 40°
Or                   ∠OAB = 40°                                        …(i)
Here, OA ⏊ AT
⟹∠OAT = 90°          
⟹       ∠OAB + ∠BAT = 90°
⟹                  ∠BAT = 90° – 40° = 50°    [Using (i)]
Hence, the value of ∠BAT is 50°.
Question. 4 From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle is drawn. Then, the area of the quadrilateral PQOR is
(a) 60 cm²         
(b) 65 cm²       
(c) 30 cm²         
(d) 32.5 cm²
Solution. (a)
Explanation:
Given conditions are described in the diagram given below:


Question. 5 At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A, is
(a) 4 cm    
(b) 5 cm    
(c) 6 cm    
(d) 8cm
Solution. (d)        
Explanation:
Given conditions are described in the diagram given below:

Now, as the perpendicular from centre to the chord of the same circle bisects that chord.
Hence, length of chord CD = 2EC = 2 × 4 = 8 cm
Exercise 9.1
Multiple Choice Questions (Q. No. 6-10)
Question. 6 In figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ÐOTA = 30°. Then, AT is equal to

Answer. (c)
Explanation:
Join OA.

Question. 7 In figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to

Answer. (a)
Explanation:
As, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴                ∠OPR = 90°
⟹∠OPQ  + ∠QPR = 90°                   
⟹∠OPQ = 90° – 50° = 40°                [Given, ∠QPR = 50°]
Now, in ΔOPQ,
                   OP = OQ = Radius of circle
⟹∠OQP = ∠OPQ = 40°                    [Angles opposite to equal sides of a triangle are also equal]                                                      
In ΔOPQ,  
                  ∠O + ∠P + ∠Q = 180°                    [Sum of angles of a triangle = 180°]
⟹∠O = 180° – (40° + 40°)                [∵ ∠P = ∠Q = 40°]
⟹∠O = 180° – 80° = 100°
Or               ∠POQ = 100°
Question. 8 In figure, if PA and PB are tangents to the circle with centre O such that ÐAPB = 50°, then ÐOAB is equal to

Answer. (a)
Explanation:
As, we know that the tangents drawn from an external point to a circle are equal
∴                  PA = PB 
Thus, in ΔPAB,
∠PBA = ∠PAB                 [Angles opposite to equal sides of a triangle are also equal]
Let ∠PBA = ∠PAB = θ
Now, in ΔPAB,     
                  ∠P + ∠A + ∠B = 180°    [Sum of angles of a triangle = 180°]
⟹               50° + θ + θ = 180°
⟹2θ = 180° – 50° = 130°
⟹               θ = 65°
Now, as we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴                  OA ⏊ PA
⟹               ∠PAO = 90°
⟹               ∠PAB + ∠OAB = 90°
⟹               65° + ∠OAB = 90°
⟹              ∠OAB = 90° – 65° = 25°
Question. 9 If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then the length of each tangent is

Answer. (d)
Explanation:
Let PQ and PR be the two tangents drawn from external point P with angle between them being 60°.

Join OQ and OR.
We know that any two tangents drawn to a circle from an external point are equally inclined to the line segment, joining the centre to that point.

Question. 10 In figure, if PQRis the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to

Answer. (b)
Explanation:

Exercise 9.2
Very Short Answer Type Questions (Q. No. 1-5)
Questions. 1 If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents at A and B is also 60°.
Solution. False
Given conditions can be expressed in the form of following diagram:

We know that a tangent at any point on a circle is perpendicular to the radius through the point of contact.

Questions. 2 The length of tangent from an external point P on a circle is always greater than the radius of the circle.
Solution. False
The length of a tangent from an external point may or may not be greater than the radius of the circle.
It can be clear from the following diagrams:

Questions. 3 The length of tangent from an external point P on a circle with centre O is always less than OP.
Solution. True
Let PT be a tangent at point T lying on a circle with centre O. Join OT.

As, in a right angled triangle, hypotenuse is always greater than any of the two sides of the triangle.
∴              OP > PT
or            PT < OP
Hence, the length of tangent from an external point P on a circle with centre O is always less than OP.
Questions. 4 The angle between two tangents to a circle may be 0°.
Solution. True
Angle between two tangents to a circle will be 0° only when the two tangents will coincide with each other or will be parallel to each other.
Questions. 5 If angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°, then.
Solution. True
Given condition can be represented as given in the following diagram:

We know that any two tangents drawn to a circle from an external point are equally inclined to the line segment, joining the centre to that point.

Exercise 9.2
Very Short Answer Type Questions (Q. No. 6-10)


Question. 7 The tangent to the circumcircle of an isosceles ΔABC at A, in which AB = AC, is parallel to BC.
Solution.   True
Let DE be the tangent to the circumcircle of an isosceles ΔABC at A.

Question. 8 If a number of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular bisector of PQ.
Solution. False
Let circles S₁, S₂, S₃, S₄, … with centres C₁, C₂, C₃, C₄, …, respectively, touch the line segment PQ at a point A.

We know that perpendicular on any point of a segment PQ may be only one. Therefore, all the line segments C₁A, C₂A. C₃A,…,so on are coincident.
But as A is not mid point of PQ therefore, perpendicular AB will not be the perpendicular bisector of PQ.  
Thus, the centre of each circle lies on a line segment which is perpendicular to PQ but not its bisector.
Question. 9 If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie
on the perpendicular bisector of PQ.
Solution. True
Let S₁, S₂ and S₃  be the circles with centres C₁, C₂ andC₃, respectively,passing through the end points P and Q of a line segment PQ.

As we know that the perpendicular bisectors of a chord of a circle always passes through the centre of circle.
Thus, perpendicular bisector of PQ passes through C₁, C₂ andC₃. Similarly, all the circle passing through PQ will have their centre on perpendiculars bisectors of PQ.
Question. 10 AB is a diameter of a circle and AC is its chord such that Ð BAC = 30°. If the tangent at C intersects A extended at D, then BC = BD.
Solution. True
Consider the following diagram:

Exercise 9.3
Short Answer Type Questions (Q. No. 1-5)
Question. 1 Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm
is a tangent to the inner circle. Find the radius of the inner circle.
Solution:
Let C₁ and C₂ be the two concentric circles with centre at O. Radius of outer circle C₂ is 5 cm. AC is a 8 cm long chord which touches the inner circle C₁ at point B.
Join OB.

Question. 2 Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.
Solution:
Consider the following diagram:

Question. 3 Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
Solution:
Given: Two tangents PQ and PR are drawn from an external point P to a circle with centre O.                                        
To prove: Centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
Construction: Join OR, and OQ.

Thus, the centre O lies on angle bisecter of PR and PQ.              
Hence proved.
Question. 4 If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO i.e., BO = 2BC.
Solution:
Given: BD and BC are two tangents drawn from an external point B such that ∠DBC = 120°
Construction: Join OC, OD and OB.

To prove: BO = 2BC
Proof:
Since, tangent at any point on a circle is perpendicular to the radius through the point of contact,

Question. 5 In figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD

Solution:
Given: AB and CD are common tangents to two circles of unequal radii.
To prove: AB = CD

Hence proved
Exercise 9.3
Short Answer Type Questions (Q. No. 6-10)
Question. 6 In figure, AB and CD are common tangents to two circles of equal radii. Prove that AB = CD.     

Solution.
Given: Two circles of equal radii, two common tangents, AB and CD on circles C₁ and C₂ with centres O₁ and O₂.
To prove: AB = CD
Construction: Join AC, BD, O₂A, O₂C, O₂B, O₂D, O₁A and O₁C.

Proof:

Question. 7 In figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.

Question. 8 A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.
Solution.
Given: Chord PQ of a circle is parallel to tangent drawn at point R of that circle.

Question. 9 Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
Solution.
Given: Two tangents are drawn at the ends of the chord of a circle.
To prove: ∠1 = ∠2
Proof:
Let two tangents drawn at the ends of a chord AB intersect at point C.

As, we know that tangents drawn from an external point to a circle are equal,
∴   AC = BC
⟹ ∠2 = ∠1           [Angles opposite to equal sides of a triangle are equal]
Hence, tangents AC and BC make equal angles with chord AB.
Hence, proved.
Question. 10 Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.
Solution.
Given: AB is a diameter of the circle with centre O and a chord CD is parallel to tangent MAN .

Now, we know that a perpendicular drawn from centre of circle to chord bisects the chord.
Thus, OE bisects CD.         
Similarly, the diameter AB bisects all the chords which are parallel to the tangent at the point A.
Exercise 9.4
Long Answer Type Questions (Q. No. 1-7)
Question. 1 If a hexagon ABCDEF circumscribe a circle, prove that
                AB + CD + EF = BC + DE + FA
Solution.
Given: A hexagon ABCDEF circumscribes a circle such that the sides, AB, BC, CD, DE and DF touches the circle at P, Q, R, S, T and U respectively.

To prove: AB + CD + EF = BC + DE+ FA
Proof:
As we know that the tangents drawn from an external point to the circle are equal,
∴   AP = AU
     BP = BQ
     CR = CQ
     DS = DR
     ES = ET
     FT = FU  ……(i)
Now, AB + CD + EF = (AP + BP) + (CR + DR) + (ET + FT)
= AU + BQ + CQ + DS + ES + FU    [Using equations in (i)]
= (AU + FU) + (BQ + CQ) + (DS + ES)
⟹  AB + CD + EF = AF + BC + DE
Hence proved.
Question. 2 Let s denotes the semi-perimeter of a ΔABC in which BC = a, CA = b and AB = c. If a circle touches the sides BC, CA, AB at D, E, F respectively, prove that BD = s – b.
Solution.
Given:
A circle is inscribed in the ΔABC, which touches BC, CA and AB at D, E and F respectively. Also BC = a, CA = b and AB = c.
And s is the semi-perimeter of a ΔABC


Question. 3 From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD.
Solution.
Given: PA and PB are two tangents drawn to a circle with centre O. A tangent CD at E, intersects AP and PB at C and D respectively.

To find: Perimeter of ΔPCD
Proof:
Since, tangents drawn from an external point to a circle are equal,
∴  CE = CA, DE = DB, PA = PB   ….(i)
Now, perimeter of DPCD = PC + CD+PD
                    = PC + CE + ED + PD
                    = PC + CA + DB + PD
                    = PA + PB
                    = 2PA = 2 ×10
                    = 20 cm
Question. 4 If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB.

Solution.
Given: Chord AB, diameter AOC and tangents AT of a circle with centre O.
To prove: ∠BAT = ∠ACB
Proof:
Here AC is a diameter and we know that diameter of a circle always subtends a right angle at any point on the circle.
∴            ∠ABC = 90°               

Question. 5 Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.
Solution.
Two circles with centres O and O’, and radii 3 cm and 4 cm respectively are drawn to intersect each other at points P and Q. OP and PO’ are two tangents drawn at point P.
Join OO’ such that it intersects PQ at R.

Since, tangent at any point of circle is perpendicular to radius through the point of contact,
∴              ∠OPO’= 90°  
Thus by using Pythagoras theorem in right angled DOPO’, we get:


Now, as the perpendicular drawn from the centre bisects the chord.
Therefore, length of common chord, PQ = 2 PR = 2 × 2.4 = 4.8cm
Question. 6 In a right angle ΔABC is which ∠B = 90°, a circle is drawn with AB as diameter intersecting the
hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.
Solution.
Given conditions can be drawn as the following diagram:


Question. 7 In figure, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn
parallel to the tangent PQ. Find the ∠RQS.

Solution.
Given, PQ and PR are tangents from an external point P to circle.

Exercise 9.4
Long Answer Type Questions (Q. No. 8-14)
Question. 8 AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects extended AB at a point D. Prove that BC = BD.
Solution.
Given: A circle with centre O and AB its diameter, AC is a chord such that ∠BAC = 30°.

To Prove: BC = BD
Proof:
We know that the angle between tangent and the chord equals angle made by the chord in alternate segment.

Question. 9 Prove that the tangent drawn at the mid point of an arc of a circle is parallel to the chord joining the end points of the arc
Solution.
Given: Consider an arc ACB in which C is mid point of arc ACB and PCQ is the tangent to the circle.
Join AC, BC and AB.


Question. 10 In a figure the common tangents, AB and CD to two circles with centres O and O’ intersect at E.
Prove that the points O, E and O’ are collinear.

Solution.
Given: AB and CD are the two common tangents to two circles with centres O and O’.
AB and CD intersect at E.
To prove: Points O, E and O’ are collinear
Construction: Joint AO, OC and O’D, O’B
Similarly, OE is the angle bisector of ÐAEC.


Question. 11 In figure, O is the centre of a circle of radius 5 cm, T is a point such that OT = 13 and OT
intersects the circle at E, if AB is the tangent to the circle at E, find the length of AB.



Question. 12 The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA =
110°, find ∠CBA.
[Hint: Join C with centre O].

But angle between tangent and the chord equals the angle made by the chord in alternate segment

Question. 13 If an isosceles ΔABC in which AB = AC = 6cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.
Solution.
Given that an isosceles DABC is inscribed in a circle with centre O  and radius 9 cm.
Join OB, OC and OA.