Rbse Solutions for Class 10 Maths Chapter 10 Exercise 10.2 | Properties of Tangents

Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 10 Exercise 10.2 on Circles. Master the application of the Radius-Tangent Theorem and the theorem that tangents from an external point are equal in length (Q.1, Q.6). Prove core circle theorems including the relationship between the angle between tangents and the angle subtended at the center (Q.2, Q.10, Q.13), and proving that tangents at the ends of a diameter are parallel (Q.4). Solutions include proofs that a circumscribing parallelogram is a rhombus (Q.11) and using the tangent property with Heron’s formula to find the sides of a circumscribed triangle (Q.12). Essential practice for geometric proofs and advanced circle problems.

This exercise uses two key theorems about tangents:

1. Radius-Tangent Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
2. Tangents from an External Point Theorem: The lengths of tangents drawn from an external point to a circle are equal.

Q.1 to 3: Choose the correct option and justify

1. Radius of the circle

Let O be the center, P be the point of contact, and Q be the external point.

• Length of tangent PQ=24 cm.
• Distance of Q from center OQ=25 cm.
• Radius OP=r (to be found).

In the right triangle △OPQ (right-angled at P):

OP2+PQ2=OQ2

r2+242=252

r2+576=625

r2=625−576=49

r=49​=7 cm

The correct option is (A) 7 cm.

2. Find ∠PTQ

TP and TQ are tangents to a circle with center O. P and Q are points of contact.

• By the Radius-Tangent Theorem, ∠OPT=90∘ and ∠OQT=90∘.
• OPTQ is a quadrilateral. The sum of angles in a quadrilateral is 360∘.∠POQ+∠OPT+∠OQT+∠PTQ=360∘110∘+90∘+90∘+∠PTQ=360∘290∘+∠PTQ=360∘∠PTQ=360∘−290∘=70∘

The correct option is (B) 70∘.

3. Find ∠POA

PA and PB are tangents from point P to a circle with center O. ∠APB=80∘.

• Consider △OAP and △OBP.
  • OA=OB (Radii of the same circle).
  • OP=OP (Common side).
  • PA=PB (Tangents from external point P).
• Therefore, △OAP≅△OBP (SSS congruence rule).
• By CPCT, ∠APO=∠BPO and ∠POA=∠POB.
• Since OP bisects ∠APB:∠APO=21​∠APB=21​(80∘)=40∘
• In the right triangle △OAP (∠OAP=90∘):∠POA+∠APO+∠OAP=180∘∠POA+40∘+90∘=180∘∠POA=180∘−130∘=50∘

The correct option is (A) 50∘.

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

• Let AB be a diameter of the circle with center O.
• Let l be the tangent at point A, and m be the tangent at point B.
• Since OA is the radius through the point of contact A, the Radius-Tangent Theorem states OA⊥l.∠OAP=90∘
• Similarly, since OB is the radius through the point of contact B, OB⊥m.∠OBQ=90∘
• Since A,O,B are collinear (forming the diameter), the line segment AB is a transversal intersecting lines l and m.
• The angles ∠OAP and ∠OBQ are alternate interior angles (or consecutive interior angles).
  • Consider the alternate interior angles formed by transversal AB: ∠OAP and ∠OBQ are not alternate interior angles. Consider transversal AB.
  • The angles ∠OAP and ∠OBR (where R is on m on the opposite side of Q) are consecutive interior angles.
  • More simply, consider the alternate interior angles formed by the transversal AB and the two lines l and m. Since ∠OAP=90∘ and ∠OBQ=90∘, we can see that:
    • If we extend l and m, we have ∠DAB=90∘ and ∠CBA=90∘.
    • ∠DAB and ∠CBA are on the same side of the transversal AB.
• The sum of interior angles on the same side of the transversal AB is ∠OAP+∠OBQ=90∘+90∘=180∘.
• Alternate Interior Angles: Consider any line P′AQ passing through A on l and R′BS passing through B on m.
  • ∠OAP=90∘. ∠OBA is part of the diameter line.
  • Since OA⊥l, ∠OAP=90∘.
  • Since OB⊥m, ∠OBQ=90∘.
  • Since A,O,B is a straight line, ∠OAP and ∠OBP (if P and Q were on the same side) are consecutive interior angles on transversal AB and lines l,m. Their sum is ∠PAB+∠QBA=90∘+90∘=180∘.
  • Therefore, the lines l and m are parallel.

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

• Let l be the tangent to the circle at point P.
• We know from the Radius-Tangent Theorem that the radius OP is perpendicular to the tangent l. That is, ∠OPA=90∘.
• Assume there is another line l′ passing through P that is perpendicular to l, but does not pass through the center O. Let l′ pass through O′.
• If l′ is perpendicular to l at P, then ∠O′PA=90∘.
• This means that both the radius OP and the line segment O′P are perpendicular to the same line l at the same point P.
• By the Uniqueness of Perpendiculars, there can be only one line perpendicular to a given line at a given point.
• Therefore, the line O′P must coincide with the line OP.
• Since the line O′P coincides with the line OP, any line perpendicular to the tangent at the point of contact must pass through the center O.

6. Find the radius of the circle.

Let O be the center, P be the point of contact, and A be the external point.

• Distance from center OA=5 cm.
• Length of tangent PA=4 cm.
• Radius OP=r (to be found).

In the right triangle △OPA (right-angled at P):

OP2+PA2=OA2

r2+42=52

r2+16=25

r2=25−16=9

r=9​=3 cm

The radius of the circle is 3 cm.

7. Find the length of the chord.

Let O be the center of two concentric circles.

• Radius of larger circle R=5 cm.
• Radius of smaller circle r=3 cm.
• Let AB be the chord of the larger circle that touches the smaller circle at point P.
• Since AB is tangent to the smaller circle at P, the radius OP is perpendicular to AB (OP⊥AB). OP=r=3 cm.
• OA is the radius of the larger circle. OA=R=5 cm.
• In the right triangle △OPA (right-angled at P):OP2+AP2=OA232+AP2=529+AP2=25AP2=16⟹AP=4 cm
• Since the perpendicular from the center to a chord bisects the chord, AB=2×AP.AB=2×4=8 cm The length of the chord is 8 cm.

8. Prove that AB+CD=AD+BC.

Let the quadrilateral ABCD circumscribe a circle, touching it at points P,Q,R,S respectively, such that P is on AB, Q on BC, R on CD, and S on DA. Using the Tangents from an External Point Theorem:

• Tangents from A: AP=AS…(1)
• Tangents from B: BP=BQ…(2)
• Tangents from C: CR=CQ…(3)
• Tangents from D: DR=DS…(4)

Add all four equations:

AP+BP+CR+DR=AS+BQ+CQ+DS

Rearrange the terms:

(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)

From the figure, we see that:

• AP+BP=AB
• CR+DR=CD
• AS+DS=AD
• BQ+CQ=BC

Substitute these back:

AB+CD=AD+BC

The sum of the opposite sides of a quadrilateral circumscribing a circle are equal.

9. Prove that ∠AOB=90∘.

• XY and X′Y′ are parallel tangents. AB is another tangent with point of contact C.
• O is the center. OA and OB connect O to the external points A and B.
• Consider the quadrilateral APQC.
  • ∠OPA=90∘ and ∠OCB=90∘.
• Consider △OPA and △OCA.
  • OP=OC (Radii).
  • OA=OA (Common side).
  • AP=AC (Tangents from external point A).
  • △OPA≅△OCA (SSS congruence).
  • By CPCT, ∠POA=∠COA. Let ∠COA=x.
• Similarly, consider △OQB and △OCB.
  • OQ=OC (Radii).
  • OB=OB (Common side).
  • QB=CB (Tangents from external point B).
  • △OQB≅△OCB (SSS congruence).
  • By CPCT, ∠QOB=∠COB. Let ∠COB=y.
• Since XY and X′Y′ are parallel, the line POQ (which is a straight line/diameter perpendicular to XY and X′Y′) forms a straight angle ∠POQ=180∘.∠POA+∠COA+∠COB+∠QOB=180∘x+x+y+y=180∘2x+2y=180∘2(x+y)=180∘x+y=90∘
• Since ∠AOB=∠COA+∠COB=x+y:∠AOB=90∘

10. Prove that the angle between two tangents is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

• Let P be the external point, PA and PB be the tangents, and A and B be the points of contact.
• We need to prove that ∠APB and ∠AOB are supplementary, i.e., ∠APB+∠AOB=180∘.
• Consider the quadrilateral OAPB. The sum of its interior angles is 360∘.∠OAP+∠APB+∠PBO+∠BOA=360∘
• By the Radius-Tangent Theorem:
  • ∠OAP=90∘ (Radius OA⊥ Tangent PA).
  • ∠PBO=90∘ (Radius OB⊥ Tangent PB).
• Substitute these values:90∘+∠APB+90∘+∠AOB=360∘180∘+∠APB+∠AOB=360∘∠APB+∠AOB=180∘ Since the sum of the angles is 180∘, the two angles are supplementary.

11. Prove that the parallelogram circumscribing a circle is a rhombus.

• Let ABCD be a parallelogram circumscribing a circle.
• Since ABCD is a parallelogram, its opposite sides are equal:AB=CD…(1)BC=AD…(2)
• From Question 8, we know that for any quadrilateral circumscribing a circle:AB+CD=AD+BC
• Substitute the parallelogram properties (1) and (2) into this equation:AB+AB=AD+AD2AB=2ADAB=AD
• Now we have AB=AD. Substitute this back into (1) and (2):AB=AD⟹AB=AD=BC=CD
• Since all four sides of the parallelogram ABCD are equal, it must be a rhombus.

12. Find the sides AB and AC.

• △ABC circumscribes a circle with radius r=4 cm.
• Point of contact on BC is D, dividing it into BD=8 cm and DC=6 cm.
• Let E and F be the points of contact on AB and AC respectively.
• Using the Tangents from an External Point Theorem:
  • CD=CF=6 cm
  • BD=BE=8 cm
  • Let AE=AF=x cm (to be found).
• The sides of △ABC:
  • a=BC=8+6=14 cm
  • c=AB=x+8
  • b=AC=x+6
• Semi-perimeter (s):s=2a+b+c​=214+(x+6)+(x+8)​=228+2x​=14+x
• Area of △ABC (using Heron’s formula):Area(△ABC)=s(s−a)(s−b)(s−c)​s−a=(14+x)−14=xs−b=(14+x)−(x+6)=8s−c=(14+x)−(x+8)=6Area=(14+x)(x)(8)(6)​=48x(14+x)​…(i)
• Area of △ABC (using Area=r⋅s): The area of a triangle circumscribing a circle is also given by Area=r⋅s, where r is the inradius.Area=4(14+x)…(ii)
• Equate (i) and (ii) and solve for x:48x(14+x)​=4(14+x) Square both sides:48x(14+x)=16(14+x)2 Divide by 16(14+x) (since 14+x=0):1648x​=14+x3x=14+x2x=14⟹x=7 cm
• Find the sides AB and AC:AB=x+8=7+8=15 cmAC=x+6=7+6=13 cm

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

• Let ABCD be the quadrilateral circumscribing the circle with center O, touching at P,Q,R,S.
• We need to prove that:
  1. ∠AOB+∠COD=180∘
  2. ∠BOC+∠AOD=180∘
• From the proof of Question 9, we established that the line joining the external point to the center bisects the angle formed by the radii to the points of contact.
  • △OAP≅△OAS⟹∠1=∠8
  • △OBP≅△OBQ⟹∠2=∠3
  • △OCQ≅△OCR⟹∠4=∠5
  • △ODR≅△ODS⟹∠6=∠7
• The sum of all angles around the center O is 360∘:∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360∘
• To prove (1): ∠AOB+∠COD=180∘∠AOB=∠1+∠2;∠COD=∠5+∠6 Substitute ∠1=∠8,∠2=∠3,∠5=∠4,∠6=∠7:2∠1+2∠2+2∠5+2∠6=360∘2(∠1+∠2)+2(∠5+∠6)=360∘(∠1+∠2)+(∠5+∠6)=180∘∠AOB+∠COD=180∘
• To prove (2): ∠BOC+∠AOD=180∘∠BOC=∠3+∠4;∠AOD=∠7+∠8 Using the same reasoning as above:2(∠3)+2(∠4)+2(∠7)+2(∠8)=360∘2(∠3+∠4)+2(∠7+∠8)=360∘(∠3+∠4)+(∠7+∠8)=180∘∠BOC+∠AOD=180∘ Hence, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.

Last Updated on November 28, 2025 by Aman Singh