RS Aggarwal Solutions Chapter 12 Circles Exercise 12A Class 10 Maths
Chapter Name  RS Aggarwal Chapter 12 Circles 
Book Name  RS Aggarwal Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 12A Solutions
1. Find the length of tangent drawn to a circle with radius 8 cm form a point 17 cm away from the center of the circle.
Solution
Let O be the center of the given circle.Let P be a point, such that
OP = 17 cm
Let OT be the radius, where
OT = 5 cm
Join TP, where TP is a tangent.
Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.
∴ OT ⊥ PT
In the right ∆OTP, we have:
OP^{2} = OT^{2} + TP^{2} [By Pythagoras’ theorem]
∴ The length of the tangent is 15 cm.
2. A point P is 25 cm away from the center of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
Solution
Draw a circle and let P be a point such that OP = 25 cm
Let TP be the tangent, so that TP = 24 cm
Join OT where OT is radius.
Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.
∴ OT ⊥ PT
In the right ∆OTP, we have:
OP^{2} = OT^{2} + TP^{2} [By Pythagoras’ theorem]
∴ The length of the tangent is 7 cm.
3. Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the larger circle which touches the smaller circle.
Solution
We know that the radius and tangent are perpendicular at their point of contactIn right triangle AOP
AO^{2} = OP^{2} + PA^{2}
⇒ (6.5)^{2} = (2.5)^{2 }+ PA^{2}
⇒ PA^{2} = 36
⇒ PA = 6 cm
Since the perpendicular drawn from the center bisects the chord.
∴ PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle is 12 cm.
4. In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F Respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the length of AD, BE and CF.
SolutionWe know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm ...(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm ....(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒ 2(AD + BD + FC) = 30
⇒ AD + BD + FC = 15 cm ...(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm
5. In the given figure, a circle touches all the four of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
SolutionLet the circle touch the sides of the quadrilateral AB, BC, CD and DA at P, Q, R and S respectively.Given, AB = 6 cm, BC = 7 cm and CD = 4 cm
Tangents drawn from an external point are equal.
∴ AP = AS, BP = BQ, CR = CQ and DR = DS
Now, AB + CD (AP + BP) + (CR + DR)
⇒ AB + CD = (AS + BQ) + (CQ + DS)
⇒ AB + CD = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ AD = (AB + CD) – BC
⇒ AD = (6 + 4) – 7
⇒ AD = 3 cm.
∴ The length of AD is 3 cm.
6. In the given figure, the chord AB of the larger of the two concentric circles, with center O, touches the smaller circle at C. Prove that AC = CB.
SolutionConstruction: Join OA, OC and OB
We know that the radius and tangent are perpendicular at their point of contact
∴ ∠OCA = ∠OCB = 90°
Now, In ∆OCA and ∆OCB
∠OCA = ∠OCB = 90°
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
∆OCA ≅ ∆OCB
∴ CA = CB
7. From an external point P, tangents PA and PB are drawn to a circle with center O. If CD is the tangent to circle at a point E and PA = 14 cm, find the perimeter of ∆PCD.
SolutionGiven, PA and PB are the tangents to a circle with center O and CD is a tangent at E and
PA = 14 cm
Tangents drawn from an external point are equal.
∴ PA = PB, CA = CE and DB = DE
Perimeter of ∆PCD = PC + CD + PD
= (PA – CA) + (CE + DE) + (PB – DB)
= (PA – CE) + (CE + DE) + (PB – DE)
= (PA + PB)
= 2PA (∵ PA = PB)
= (2 × 14) cm
= 28 cm
∴ Perimeter of ∆PCD = 28 cm
8. A circle is inscribed in a ∆ABC touching AB, BC and AC at P, Q and R respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.
SolutionGiven, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle
Tangents drawn to a circle from an external point are equal.
∴ AP = AR = 7 cm, CQ = CR = 5 cm
Now, BP = (AB – AP)
= (10 – 7)
= 3 cm
∴ BP = BQ = 3 cm
∴ BC = (BQ + QC)
⇒ BC = 3 + 5
⇒ BC = 8
∴ The length of BC is 8 cm.
9. In the given figure, PA and PB are the tangents segments to a circle with centre O. Show that the points A, O, B and P are concyclic.
SolutionHere, OA = OB
And OA ⊥ AP, OA ⊥ BP (Since tangents drawn from an external point are perpendicular to the radius at the point of contact)
∴ ∠OAP = 90°, ∠OBP = 90°
∴ ∠OAP + ∠OBP = 90° + 90° = 180°
∴ ∠AOB + ∠APB = 180° (Since, ∠OAP + ∠OBP + ∠AOB + ∠APB = 360°)
Sum of opposite angle of a quadrilateral is 180°.
Hence, A, O, B and P are concyclic.
10. In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that point of contact P bisects the base BC.
SolutionWe know that tangent segments to a circle from the same external points are congruent
Now, we have
AR = AO, BR = BP and CP = CQ
Now, AB = AC
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + OC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.
11. In the given figure, O is the centre of the two concentric circles of radii 4 cm and 6 cm respectively. AP and PB are tangents to the outer and inner circle respectively. If PA = 10 cm, find the length of PB upto one place of the decimal.
SolutionGiven, O is the center of two concentric circles of radii OA = 6 cm and OB = 4 cm.
PA and PB are the two tangents to the outer and inner circles respectively and PA = 10 cm.
Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.
∴ ∠OAP = ∠OBP = 90°
∴ From rightangled ∆OAP, OP^{2 }= OA^{2} + PA^{2}
∴ The length of PB is 10.9 cm.
12. In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BC and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ∆ABC = 54 cm^{2} then find the length of sides AB and AC.
SolutionConstruction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F.
We know that tangent segments to a circle from me same external point are congruent
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
Area (∆ABC) = Area (∆BOC) + Area(∆AOB) + Area (∆AOC)
⇒ 54 = 1/2 × BC × OD + 1/2 × AB × OE + 1/2 × AC × OF
⇒ 108 = 15 × 3 + (6 + x) × 3 + (9 + x) × 3
⇒ 36 = 15 + 6 + x + 9 + x
⇒ 36 = 30 + 2x
⇒ 2x = 6
⇒ x = 3 cm
∴ AB = 6 + 3 = 9 cm and AC = 9 + 3 = 12 cm
13. PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.
SolutionLet TR = y and TP = xWe know that the perpendicular drawn from the center to me chord bisects it.
∴ PR = RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pythagoras theorem, we have
PO^{2} = OR^{2} + PR^{2 }
⇒ 3^{2} = OR^{2} + (2.4)^{2}
⇒ OR^{2} = 3.24
⇒ OR = 1.8
Now, in right triangle TPR
By using Pythagoras theorem, we have
TP^{2} = TR^{2} + PR^{2}
⇒ x^{2} = y^{2 }+ (2.4)^{2}
⇒ x^{2} = y^{2 }+ 5.76 ...(1)
Again, In right triangle TPQ
By using Pythagoras theorem, we have
TO^{2} = TP^{2 }+ PO^{2}
⇒ (y + 1.8)^{2} = x^{2} + 32
⇒ y^{2} + 3.6y + 3.24 = x^{2} + 9
⇒ y^{2 }+ 3.6y = x^{2} + 5.76 ...(2)
Solving (1) and (2), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm
14. Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.
Solution
Suppose CD and AB are two parallel tangents of a circle with centre OConstruction: Draw a line parallel to CD passing through O i.e., OP
We know that the radius and tangent are perpendicular at their point of contact.
∠OQC = ∠ORA = 90°
Now, ∠OQC + ∠POQ = 180° (cointerior angles)
⇒ ∠POQ = 180°  90° = 90°
Similarly, Now, ∠ORA + ∠POR = 180° (cointerior angles)
⇒ ∠POQ = 180°  90° = 90°
Now, ∠POR + POQ = 90° + 90° = 180°
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180°
Hence, QR is a straight line passing through center O.
15. In the given figure, with center O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm then find the radius of the circle.
SolutionWe know that tangent segments to a circle from the same external point are congruent
Now, we have
DS = DR, AR = AQ
Now AD = 23 cm
⇒ AR + RD = 23
⇒ AR = 23 – RD
⇒ AR = 23 – 5 [∴ DS = DR = 5]
⇒ AR = 18 cm
Again. AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 – AQ
⇒ QB = 29 – 18 [∵ AR = AQ = 18]
⇒ QB = 11 cm
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
16. In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA : AT = 2 : 1.
SolutionAB is the chord passing through the center
So, AB is the diameter
Since, angle in a semicircle is a right angle
∴ ∠APB = 90°
By using alternate segment theorem
We have ∠APB = ∠PAT = 30°
Now, in ∆APB
∠BAP + ∠APB + ∠BAP = 180° (Angle sum property of triangle)
⇒ ∠BAP = 180°  90°  30° = 60°
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60° = 30° + ∠PTA
⇒ ∠PTA = 60°  30° = 30°
We know that sides opposite to equal angles are equal
∴ AP = AT
In right triangle ABP
sin ∠ABP = AP/BA
⇒ sin 30° = AT/BA
⇒ 1/2 = AT/BA °
∴ BA : AT = 2 : 1