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NCERT Class 10 Mathematics Exercise 10.2 Solutions: Circles

Step-by-step working for every question of Exercise 10.2 in NCERT's Class 10 Mathematics book, chapter “Circles”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 10.2 Q1

   The tangent from Q to a circle is 24 cm and Q is 25 cm from the centre. The radius is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm.

   1. Let O be the centre of the circle and Q be the external point.
   2. Let P be the point of contact of the tangent from Q to the circle.
   3. The radius OP is perpendicular to the tangent PQ at the point of contact.
      why?

      The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   4. Therefore, triangle OPQ is a right-angled triangle with the right angle at P.
   5. We are given that the length of the tangent PQ is 24 cm.
   6. We are given that the distance from the centre to the external point, OQ, is 25 cm.
   7. Using the Pythagoras theorem in triangle OPQ: OP² + PQ² = OQ².
      why?

      In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

   8. Substitute the known values: OP² + (24)² = (25)².
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   9. Calculate the squares: OP² + 576 = 625.
   10. Solve for OP²: OP² = 625 − 576.
   11. OP² = 49.
   12. Take the square root: OP = √49 = 7 cm.

   Answer: (A) 7 cm

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2. Ex 10.2 Q2

   TP and TQ are tangents to a circle with centre O and ∠POQ = 110° (Fig. 10.11). ∠PTQ is (A) 60° (B) 70° (C) 80° (D) 90°.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Since TP and TQ are tangents to the circle, the radii OP and OQ are perpendicular to them at the points of contact.
      why?

      The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   2. Therefore, ∠OPT = 90° and ∠OQT = 90°.
   3. Consider the quadrilateral OPTQ. The sum of its interior angles is 360°.
   4. So, ∠OPT + ∠POQ + ∠OQT + ∠PTQ = 360°.
   5. Substituting the known values: 90° + 110° + 90° + ∠PTQ = 360°.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   6. Simplifying the equation gives 290° + ∠PTQ = 360°.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   7. Solving for ∠PTQ, we get ∠PTQ = 360° − 290° = 70°.

   Answer: 70° (B)

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3. Ex 10.2 Q3

   Tangents PA and PB from P to a circle with centre O meet at 80°. ∠POA is (A) 50° (B) 60° (C) 70° (D) 80°.

   1. Given that PA and PB are tangents to the circle from an external point P, and ∠APB = 80°.
   2. The line joining the centre O to the point of contact A is perpendicular to the tangent PA, so ∠OAP = 90°.
      why?

      The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   3. In triangle OAP, the sum of angles is 180°, so ∠POA + ∠OAP + ∠OPA = 180°.
   4. The line OP bisects the angle between the tangents, so ∠OPA = 80° ÷ 2 = 40°.
   5. Substituting the values into the angle sum equation: ∠POA + 90° + 40° = 180°.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   6. Solving for ∠POA gives ∠POA = 180° − 130° = 50°.

   Answer: 50° (A)

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4. Ex 10.2 Q4

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

   1. Let AB be a diameter of the circle with centre O.
   2. Let the tangent at A be line l and the tangent at B be line m.
   3. A tangent to a circle is perpendicular to the radius at the point of contact.
      why?

      The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   4. Therefore, OA is perpendicular to l and OB is perpendicular to m.
   5. This means the angle between OA and l is 90° and the angle between OB and m is 90°.
   6. Since AB is a straight line, the alternate interior angles formed by transversal AB with lines l and m are equal (both 90°).
   7. If alternate interior angles are equal, the lines are parallel.
   8. Hence, the tangents at the ends of the diameter are parallel.

   Answer: The tangents at the ends of a diameter of a circle are parallel.

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5. Ex 10.2 Q5

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

   1. Let O be the centre of a circle and let a line l be tangent to the circle at point A.
   2. We need to prove that the perpendicular to l at A passes through O.
   3. Assume, for the sake of contradiction, that the perpendicular to l at A does not pass through O.
      why?

      Proof by contradiction: assume the opposite of what is to be proved and show that it leads to something impossible; then the assumption must be false.

   4. Let B be the foot of the perpendicular drawn from O to the tangent line l.
      why?

      The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   5. Since B is the foot of the perpendicular from O to l, OB is the shortest distance from the centre O to the line l.
   6. In the right-angled triangle OAB, OA is the hypotenuse and OB is a leg.
      why?

      In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

   7. Therefore, OA is greater than OB, i.e., OA > OB.
   8. Since OA is the radius of the circle, this means the distance from the centre to the line l (which is OB) is less than the radius.
   9. If the distance from the centre to a line is less than the radius, the line must intersect the circle at two distinct points.
   10. This implies that l is a secant, not a tangent.
   11. This contradicts the given statement that l is a tangent to the circle at A.
       why?

       Proof by contradiction: assume the opposite of what is to be proved and show that it leads to something impossible; then the assumption must be false.

   12. Hence, our assumption is wrong, and the perpendicular to the tangent at A must pass through the centre O.
       why?

       The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   Answer: Proved that the perpendicular to a tangent at its point of contact passes through the centre.

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6. Ex 10.2 Q6

   The tangent from a point A, 5 cm from the centre of a circle, is 4 cm long. Find the radius.

   1. Let O be the centre of the circle and A be the external point.
   2. Let P be the point of contact of the tangent from A.
   3. Join OP and OA.
   4. The radius is perpendicular to the tangent at the point of contact, so ∠OPA = 90°.
      why?

      The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   5. In right-angled triangle OPA, OA is the hypotenuse.
      why?

      In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

   6. Given OA = 5 cm and AP = 4 cm.
   7. By Pythagoras theorem, OP² + AP² = OA².
      why?

      In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

   8. Substitute the values: OP² + 4² = 5².
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   9. OP² + 16 = 25.
   10. OP² = 25 − 16.
   11. OP² = 9.
   12. OP = √9 = 3 cm.
   13. The radius of the circle is OP.

   Answer: 3 cm

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7. Ex 10.2 Q7

   Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that touches the smaller one.

   1. Let O be the common centre of the two concentric circles.
   2. Let AB be the chord of the larger circle that touches the smaller circle at point P.
   3. Since AB is a tangent to the smaller circle, the radius OP is perpendicular to AB.
      why?

      The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   4. In a circle, a radius perpendicular to a chord bisects the chord, so AP = PB.
      why?

      The perpendicular from the centre of a circle to a chord bisects the chord (the two right triangles it makes are congruent by RHS).

   5. In the right-angled triangle OPA, OA is the radius of the larger circle, so OA = 5 cm.
   6. OP is the radius of the smaller circle, so OP = 3 cm.
   7. Using the Pythagoras theorem in triangle OPA: AP² + OP² = OA².
      why?

      In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

   8. Substituting the values: AP² + 3² = 5².
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   9. AP² + 9 = 25.
   10. AP² = 25 − 9 = 16.
   11. AP = √16 = 4 cm.
   12. The length of the chord AB is 2 × AP = 2 × 4 = 8 cm.

   Answer: 8 cm

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8. Ex 10.2 Q8

   Quadrilateral ABCD circumscribes a circle (Fig. 10.12). Prove AB + CD = AD + BC.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Let the circle touch sides AB, BC, CD, and DA at points P, Q, R, and S respectively.
   2. By the theorem of tangents from an external point, the lengths of tangents from a point to a circle are equal.
      why?

      The two tangents drawn to a circle from a point outside it are equal in length.

   3. Therefore, AP = AS, BP = BQ, CQ = CR, and DR = DS.
   4. Add the lengths of the opposite sides AB and CD: AB + CD = (AP + PB) + (CR + RD).
   5. Substitute the equal tangent lengths into the expression: AB + CD = (AS + BQ) + (CQ + DS).
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   6. Rearrange the terms to group the tangents from the same vertices: AB + CD = (AS + DS) + (BQ + CQ).
   7. Recognize that AS + DS equals the side AD and BQ + CQ equals the side BC.
   8. Thus, AB + CD = AD + BC.

   Answer: AB + CD = AD + BC

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9. Ex 10.2 Q9

   XY and X′Y′ are parallel tangents to a circle with centre O, and a tangent AB touching at C meets XY at A and X′Y′ at B (Fig. 10.13). Prove ∠AOB = 90°.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Let XY touch the circle at P and X′Y′ at Q. Join OP, OQ, OC, OA and OB. Since XY ∥ X′Y′ and both radii are perpendicular to them, P, O, Q lie on one straight line (a diameter).
   2. In △OPA and △OCA: OP = OC (radii), OA is common and ∠OPA = ∠OCA = 90° (radius ⊥ tangent). So △OPA ≅ △OCA (RHS), which gives ∠POA = ∠COA.
      why?

      The tangent at any point of a circle is perpendicular to the radius through the point of contact.

   3. In the same way, △OQB ≅ △OCB (RHS), which gives ∠QOB = ∠COB.
   4. POQ is a straight line, so ∠POA + ∠COA + ∠COB + ∠QOB = 180°.
   5. Using the equal pairs: 2∠COA + 2∠COB = 180°, so ∠COA + ∠COB = 90°.
   6. ∠AOB = ∠COA + ∠COB = 90°.

   Answer: ∠AOB = 90°

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10. Ex 10.2 Q10

    Prove that the angle between two tangents from an external point is supplementary to the angle the chord of contact subtends at the centre.

    1. Let P be an external point from which tangents PA and PB are drawn to a circle with centre O.
    2. Join OA, OB, and OP to form the quadrilateral OAPB.
    3. Since the radius is perpendicular to the tangent at the point of contact, angle OAP and angle OBP are both 90°.
       why?

       The tangent at any point of a circle is perpendicular to the radius through the point of contact.

    4. The sum of interior angles in quadrilateral OAPB is 360°, so angle AOB + angle APB + 90° + 90° = 360°.
    5. Simplifying the equation gives angle AOB + angle APB = 180°.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    6. Thus, the angle between the tangents (angle APB) and the angle subtended by the chord of contact at the centre (angle AOB) are supplementary.

    Answer: Proved that angle APB + angle AOB = 180°

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11. Ex 10.2 Q11

    Prove that a parallelogram circumscribing a circle is a rhombus.

    1. Let ABCD be a parallelogram circumscribing a circle with centre O.
    2. Let the circle touch the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.
    3. We know that the lengths of tangents drawn from an external point to a circle are equal.
       why?

       The two tangents drawn to a circle from a point outside it are equal in length.

    4. Therefore, AP = AS, BP = BQ, CQ = CR, and DR = DS.
    5. Let AP = AS = a, BP = BQ = b, CQ = CR = c, and DR = DS = d.
    6. The side AB is the sum of AP and BP, so AB = a + b.
    7. The side BC is the sum of BQ and CQ, so BC = b + c.
    8. The side CD is the sum of CR and DR, so CD = c + d.
    9. The side DA is the sum of DS and AS, so DA = d + a.
    10. Since ABCD is a parallelogram, opposite sides are equal, so AB = CD and BC = DA.
    11. Equating AB and CD gives a + b = c + d.
    12. Equating BC and DA gives b + c = d + a.
    13. From the equation b + c = d + a, we can rearrange it to get b - a = d - c.
    14. Substituting this value of (d - c) into the first equation a + b = c + d, we get a + b = c + (b - a + c).
        why?

        Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    15. Simplifying this gives a + b = b + 2c - a, which leads to 2a = 2c, so a = c.
        why?

        Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    16. Since a = c, substituting back into b - a = d - c gives b - a = d - a, so b = d.
        why?

        Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    17. Now, AB = a + b and BC = b + c. Since a = c, AB = a + b and BC = b + a.
    18. Thus, AB = BC.
    19. Since adjacent sides of the parallelogram are equal, all four sides are equal.
    20. A parallelogram with all sides equal is a rhombus.

    Answer: Proved that a parallelogram circumscribing a circle is a rhombus.

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12. Ex 10.2 Q12

    △ABC circumscribes a circle of radius 4 cm; the point of contact D splits BC into BD = 8 cm and DC = 6 cm (Fig. 10.14). Find AB and AC.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Let the circle touch sides AB, AC, and BC at points E, F, and D respectively.
    2. Since tangents drawn from an external point to a circle are equal, AE = AF, BE = BD, and CF = CD.
       why?

       The two tangents drawn to a circle from a point outside it are equal in length.

    3. Given BD = 8 cm and DC = 6 cm, we have BE = 8 cm and CF = 6 cm.
       why?

       For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

    4. Let AE = AF = x cm. Then AB = AE + BE = x + 8 and AC = AF + CF = x + 6.
       why?

       For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

    5. The area of triangle ABC is the sum of the areas of triangles OAB, OBC, and OCA, where O is the center of the circle.
    6. Area of ABC = (1/2) × r × (AB + BC + CA) = (1/2) × 4 × ((x + 8) + (8 + 6) + (x + 6)) = 2(2x + 28) = 4x + 56.
    7. Using Heron's formula, s = (AB + BC + CA) / 2 = (2x + 28) / 2 = x + 14.
    8. Area of ABC = √[s(s - AB)(s - BC)(s - CA)] = √[(x + 14)(x + 14 - x - 8)(x + 14 - 14)(x + 14 - x - 6)].
    9. Simplifying the terms inside the square root: √[(x + 14)(6)(x)(8)] = √[48x(x + 14)].
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    10. Equating the two expressions for the area: 4x + 56 = √[48x(x + 14)].
    11. Square both sides: (4x + 56)² = 48x(x + 14).
        why?

        Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    12. 16(x + 14)² = 48x(x + 14).
    13. Divide both sides by 16(x + 14) (since x ≠ -14): x + 14 = 3x.
        why?

        Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    14. Solving for x: 2x = 14, so x = 7.
    15. Therefore, AB = x + 8 = 7 + 8 = 15 cm and AC = x + 6 = 7 + 6 = 13 cm.

    Answer: AB = 15 cm, AC = 13 cm

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13. Ex 10.2 Q13

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

    1. Let ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S. Join OA, OB, OC, OD and OP, OQ, OR, OS.
    2. In △OAP and △OAS: OP = OS (radii), OA is common and AP = AS (tangents from A). So they are congruent (SSS) and ∠AOP = ∠AOS. Call this angle ∠1.
       why?

       The tangent at any point of a circle is perpendicular to the radius through the point of contact.

    3. In the same way, the tangents from B, C and D give ∠BOP = ∠BOQ (∠2), ∠COQ = ∠COR (∠3) and ∠DOR = ∠DOS (∠4).
    4. The angles around O add up to 360°: 2∠1 + 2∠2 + 2∠3 + 2∠4 = 360°, so ∠1 + ∠2 + ∠3 + ∠4 = 180°.
    5. Side AB subtends ∠AOB = ∠1 + ∠2 and the opposite side CD subtends ∠COD = ∠3 + ∠4, so ∠AOB + ∠COD = 180°.
    6. In the same way, ∠BOC + ∠AOD = (∠2 + ∠3) + (∠1 + ∠4) = 180°. Opposite sides subtend supplementary angles at the centre.

    Answer: ∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°: proved.

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