Q.1. In the given figure, O is the center of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then, ∠POQ = ?
(a) 130°
(b) 100°
(c) 90°
(d) 75°
In the given figure PT is a tangent to circle ∴ we have
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPQ + ∠QPT = 90°
∠OPQ + 50° = 90°
∠OPQ = 40°
Now, In △POQ
OP = OQ
∠PQO = ∠QPO = 40°
[Angles opposite to equal sides are equal]
Now,
∠ PQO + ∠QPO + ∠ POQ = 180° [By angle sum property of triangle]
40° + 40° + ∠POQ = 180°
∠POQ = 100°
Q.2. If the angle between two radii of a circle is 130° then the angle between the tangents at the ends of the radii is
(a) 65°
(b) 40°
(c) 50°
(d) 90°
Let us consider a circle with center O and OA and OB are two radii such that ∠AOB = 60° .
AP and BP are two intersecting tangents at point P at point A and B respectively on the circle .
To find : Angle between tangents, i.e. ∠APB
As AP and BP are tangents to given circle,
We have,
OA ⏊ AP and OB ⏊ BP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP,
By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
90° + 90° + ∠APB + 130° = 360°
∠APB = 50°
Q.3. If tangents PA and PB from a point P to a circle with center O are drawn so that ∠APB = 80° then ∠POA = ?
(a) 40°
(b) 50°
(c) 80°
(d) 60°
In △AOP and △BOP
AP = BP
[Tangents drawn from an external point are equal]
OP = OP [Common]
OA = OB
[Radii of same circle]
△AOP ≅ △BOP
[By Side  Side  Side criterion]
∠APO = ∠BPO
[Corresponding parts of congruent triangles are congruent]
∠APB = ∠APO + ∠BPO
80 = ∠APO + ∠APO
2∠APO = 80
∠APO = 40°
In △AOP
∠APO + ∠AOP + ∠OAP = 180°
[By angle sum property]
40° + ∠AOP + 90° = 180°
[ ∠OAP = 90° as OA ⏊ AP because Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠AOP = 50°
Q.4. In the given figure, AD and AE are the tangents to a circle with center O and BC touches the circle at F. If AE = 5 cm then perimeter of ΔABC is
(a) 15 cm
(b) 10 cm
(c) 22.5 cm
(d) 20 cm
Given : From an external point A, two tangents, AD and AE are drawn to a circle with center O. At a point F on the circle tangent is drawn which intersects AE and AD at B and C, respectively. And AE = 5 cm
To Find : Perimeter of △ABC
As we know that, Tangents drawn from an external point to a circle are equal.
So we have
BE = BF …[1]
[Tangents from point B]
CF = CD …[2]
[Tangents from point C]
Now Perimeter of Triangle abc
= AB + BC + AC
= AB + BF + CF + AC
= AB + BE + CD + AC …[From 1 and 2]
= AE + AD
Now,
AE = AD = 5 cm as tangents drawn from an external point to a circle are equal
So we have
Perimeter = AE + AD = 5 + 5 = 10 cm
Q.5. In the given figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively.
If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, find x.
As we know, Tangents drawn from an external point are equal.
CR = CQ [tangents from point C]
CQ = 3 cm [as CR = 3 cm]
Also,
BC = BQ + CQ
7 = BQ + 3 [BC = 7 cm]
BQ = 4 cm
Now,
BP = BQ [tangents from point B]
BP = 4 cm …[1]
AP = AS [tangents from point A]
AP = 5 cm [As AC = 5 cm] ….[2]
AB = AP + BP = 5 + 4 = 9 cm [From 1 and 2]
AB = x = 9 cm
Q.6. In the given figure, PA and PB are the tangents to a circle with center O. Show that the points A, 0, B, P are concyclic.
Given : PA and PB are tangents to a circle with center O
To show : A, O, B and P are concyclic i.e. they lie on a circle i.e. AOBP is a cyclic quadrilateral.
Proof :
OB ⏊ PB and OA ⏊ AP
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OBP = ∠OAP = 90°
∠OBP + ∠OAP = 90 + 90 = 180°
AOBP is a cyclic quadrilateral i.e. A, O, B and P are concyclic.
[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic ]
Hence Proved.
Q.7. In the given figure, PA and PB are two tangents from an external point P to a circle with center O. If ∠PBA = 65°, find ∠OAB and ∠APB.
In the given Figure,
PA = PB
[Tangents drawn from an external points are equal]
∠PBA = ∠PAB
[Angles opposite to equal sides are equal]
∠PBA = ∠PAB = 65°
In △APB
∠PAB + ∠PBA + ∠APB = 180°
65° + 65° + ∠APB = 180°
∠APB = 50°
Also,
OB ⏊ AP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAP = 90°
∠OAB + ∠PAB = 90°
∠OAB + 65° = 90°
∠OAB = 25°
Q.8. Two tangent segments BC and BD are drawn to a circle with center O such that ∠CBD = 120°. Prove that OB = 2BC.
Given : A circle with center O , BC and BD are two tangents such that ∠CBD = 120°
To Proof : OB = 2BC
Proof :
In △BOC and △BOD
BC = BD
[Tangents drawn from an external point are equal]
OB = OB
[Common]
OC = OD
[Radii of same circle]
△BOC ≅ △BOD [By Side  Side  Side criterion]
∠OBC = ∠OBD
[Corresponding parts of congruent triangles are congruent]
∠OBC + ∠OBD = ∠CBD
∠OBC + ∠OBC = 120°
2 ∠OBC = 120°
∠OBC = 60°
In △OBC
⇒ OB = 2BC
Hence Proved !
Q.9. Fill in the blanks.
(i) A line intersecting a circle in two distinct points is called a ……..
(ii) A circle can have ………..parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the ………..
(iv) A circle can have ………..tangents.
(i) secant
(ii) two
(iii) point of contact
(iv) infinitely many
Q.10. Prove that the lengths of two tangents drawn from an external point to a circle are equal.
Let us consider a circle with center O.
TP and TQ are two tangents from point T to the circle.
To Proof : PT = QT
Proof :
OP ⏊ PT and OQ ⏊ QT
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPT = ∠OQT = 90°
In △TOP and △QOT
∠OPT = ∠OQT
[Both 90°]
OP = OQ
[Common]
OT = OT
[Radii of same circle]
△TOP ≅ △QOT
[By Right Angle  Hypotenuse  Side criterion]
PT = QT
[Corresponding parts of congruent triangles are congruent]
Hence Proved.
Q.11. Prove that the tangents drawn at the ends of the diameter of a circle are parallel.
Let AB be the diameter of a circle with center O.
CD and EF are two tangents at ends A and B respectively.
To Prove : CD  EF
Proof :
OA ⏊ CD and OB ⏊ EF
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAD = ∠OBE = 90°
∠OAD + ∠OBE = 90° + 90° = 180°
Considering AB as a transversal
⇒ CD  EF
[Two sides are parallel, if any pair of the interior angles on the same sides of transversal is supplementary]
Q.12. In the given figure, if AB = AC, prove that BE = CE.
We know, that tangents drawn from an external point are equal.
AD = AF
[tangents from point A] [1]
BD = BE
[tangents from point B] [2]
CF = CE
[tangents from point C] [3]
Now,
AB = AC [Given] …[4]
Substracting [1] From [4]
AB  AD = AC  AF
BD = CF
BE = CE [From 2 and 3]
Hence Proved.
Q.13. If two tangents are drawn to a circle from an external point, show that they subtend equal angles at the center.
Let PT and PQ are two tangents from external point P to a circle with center O
To Prove : PT and PQ subtends equal angles at center i.e. ∠POT = ∠QOT
In △OPT and △OQT
OP = OQ [radii of same circle]
OT = OT [common]
PT = PQ [Tangents drawn from an external point are equal]
△OPT ≅ △OQT [By Side  Side  Side Criterion]
∠POT = ∠QOT [Corresponding parts of congruent triangles are congruent]
Hence, Proved.
Q.14. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
Let us consider a circle with center O and BC be a chord, and AB and AC are tangents drawn at end of a chord
To Prove : AB and AC make equal angles with chord, i.e. ∠ABC = ∠ACB
Proof :
In △ABC
AB = PC
[Tangents drawn from an external point to a circle are equal]
∠ACB = ∠ABC
[Angles opposite to equal sides are equal]
Hence Proved.
Q.15. Prove that the parallelogram circumscribing a circle, is a rhombus.
Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.
To Proof : ABCD is a rhombus.
As ABCD is a parallelogram
AB = CD and BC = AD …[1]
[opposite sides of a parallelogram are equal]
Now, As tangents drawn from an external point are equal.
We have
AP = AS
[tangents from point A]
BP = BQ
[tangents from point B]
CR = CQ
[tangents from point C]
DR = DS
[tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
AB + AB = BC + BC [From 1]
AB = BC …[2]
From [1] and [2]
AB = BC = CD = AD
And we know,
A parallelogram with all sides equal is a rhombus
So, ABCD is a rhombus.
Hence Proved.
Q.16. Two concentric circles are of radii 5 cm and 3 cm respectively. Find the length of the chord of the larger circle which touches the smaller circle.
Given : Two concentric circles (say C_{1} and C_{2}) with common center as O and radius r_{1} = 5 cm and r_{2} = 3 cm respectively.
To Find : Length of the chord of the larger circle which touches the circle C_{2}. i.e. Length of AB.
As AB is tangent to circle C_{2} and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"
So, we have,
OP ⏊ AB
∴ OPB is a right  angled triangle at P,
By Pythagoras Theorem in △OPB
[i.e. (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2} ]
We have,
(OP)^{2} + (PB)^{2} = (OB)^{2}
r_{2}^{2} + (PB)^{2} = r_{1}^{2}
(3)^{2} + (PB)^{2}= (5)^{2}
9 + (PB)^{2} = 25
(PB)^{2} = 16
PB = 4 cm
Now, AP = PB ,
[as perpendicular from center to chord bisects the chord and OP ⏊ AB]
So, AB = AP + PB = PB + PB = 2PB = 2(4) = 8 cm
Q.17. A quadrilateral is drawn to circumscribe a circle. Prove that the sums of opposite sides are equal.
Let us consider a quadrilateral ABCD, And a circle is circumscribed by ABCD
Also, Sides AB, BC, CD and DA touch circle at P, Q, R and S respectively.
To Proof : Sum of opposite sides are equal, i.e. AB + CD = AD + BC
Proof :
In the Figure,
As tangents drawn from an external point are equal.
We have
AP = AS
[tangents from point A]
BP = BQ
[tangents from point B]
CR = CQ
[tangents from point C]
DR = DS
[tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence Proved.
Q.18. Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
Consider a quadrilateral, ABCD circumscribing a circle with center O and AB, BC, CD and AD touch the circles at point P, Q, R and S respectively.
Joined OP, OQ, OR and OS and renamed the angles (as in diagram)
To Prove : Opposite sides subtends supplementary angles at center i.e.
∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°
Proof :
In △AOP and △AOS
AP = AS
[Tangents drawn from an external point are equal]
AO = AO
[Common]
OP = OS
[Radii of same circle]
△AOP ≅ △AOS
[By Side  Side  Side Criterion]
∠AOP = ∠AOS
[Corresponding parts of congruent triangles are congruent]
∠1 = ∠2 …[1]
Similarly, We can Prove
∠3 = ∠4 ….[2]
∠5 = ∠6 ….[3]
∠7 = ∠8 ….[4]
Now,
As the angle around a point is 360°
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
∠2 + ∠2 + ∠3 + ∠3 + ∠6 + ∠6 + ∠7 + ∠7 = 360° [From 1, 2, 3 and 4]
2(∠2 + ∠3 + ∠6 + ∠7) = 360°
∠AOB + ∠COD = 180°
[As, ∠2 + ∠3 = ∠AOB and ∠5 + ∠6 = ∠COD] [5]
Also,
∠AOB + ∠BOC + ∠COD + ∠AOD = 360°
[Angle around a point is 360°]
∠AOB + ∠COD + ∠BOC + ∠AOD = 360°
180° + ∠BOC + ∠AOD = 360° [From 5]
∠BOC + ∠AOD = 180°
Hence Proved
Q.19. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the center.
Let us consider a circle with center O and PA and PB are two tangents to the circle from an external point P
To Prove : Angle between two tangents is supplementary to the angle subtended by the line segments joining the points of contact at center, i.e. ∠APB + ∠AOB = 180°
Proof :
As AP and BP are tangents to given circle,
We have,
OA ⏊ AP and OB ⏊ BP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP, By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠AOB + ∠APB = 360°
90° + 90° + ∠AOB + ∠APB = 360°
∠AOB + ∠APB = 180°
Hence Proved.
Q.20. PQ is a chord of length 16 cm of a circle of radius 10 cm. The tangents at P and Q intersect at a point T as shown in the figure.
Find the length of TP.
Given : A circle with center O and radius 3 cm and PQ is a chord of length 4.8 cm. The tangents at P and Q intersect at point T
To Find : Length of PT
Construction : Join OQ
Now in △OPT and △OQT
OP = OQ
[radii of same circle]
PT = PQ
[tangents drawn from an external point to a circle are equal]
OT = OT
[Common]
△OPT ≅ △OQT
[By Side  Side  Side Criterion]
∠POT = ∠OQT
[Corresponding parts of congruent triangles are congruent]
or ∠POR = ∠OQR
Now in △OPR and △OQR
OP = OQ
[radii of same circle]
OR = OR [Common]
∠POR = ∠OQR [Proved Above]
△OPR ≅ △OQT
[By Side  Angle  Side Criterion]
∠ORP = ∠ORQ
[Corresponding parts of congruent triangles are congruent]
Now,
∠ORP + ∠ORQ = 180°
[Linear Pair]
∠ORP + ∠ORP = 180°
∠ORP = 90°
⇒ OR ⏊ PQ
⇒ RT ⏊ PQ
As OR ⏊ PQ and Perpendicular from center to a chord bisects the chord we have
PR = QR = PQ/2 = 16/2 = 8 cm
∴ In right  angled △OPR,
By Pythagoras Theorem
[i.e. (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2} ]
(OP)^{2 }= (OR)^{2} + (PR)^{2}
(10)^{2 }= (OR)^{2} + (8)^{2}
100 = (OR)^{2 }+ 64
(OR)^{2}= 36
OR = 6 cm
Now,
In right angled △TPR, By Pythagoras Theorem
(PT)^{2} = (PR)^{2 }+ (TR)^{2} [1]
Also, OP ⏊ OT
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
In right angled △OPT, By Pythagoras Theorem
(PT)^{2} + (OP)^{2} = (OT)^{2}
(PR)^{2} + (TR)^{2} + (OP)^{2}= (TR + OR)^{2} [From 1]
(8)^{2} + (TR)^{2} + (10)^{2} = (TR + 6)^{2}
64 + (TR)^{2} + 100 = (TR)^{2} + 2(6)TR + (6)^{2}
164 = 12TR + 36
12TR = 128
TR = 10.7 cm [Appx]
Using this in [1]
PT^{2} = (8)^{2 }+ (10.7)^{2}
PT^{2} = 64 + 114.49
PT^{2} = 178.49
PT = 13.67 cm [Appx]
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