Exercise 10.2 answers: Circles

Open NCERT p. 151Checked by computer

1. The radius OP to the point of contact P is perpendicular to the tangent PQ, so triangle OPQ is right-angled at P.
2. OQ (25 cm) is the hypotenuse, and PQ (24 cm) is one leg. By Pythagoras: OQ² = OP² + PQ².
3. 25² = OP² + 24², so OP² = 625 − 576 = 49, giving OP = 7 cm.

Answerradius = 7 cm (option A).

Watch this explained “The length without drawing anything”, 7:21 into None, one or two, depending on where the point sits

Open NCERT p. 151Checked by computer

1. TPOQ is a quadrilateral formed by the centre O, the external point T, and the two points of contact P and Q.
2. The radius meets each tangent at a right angle, so ∠OPT = 90° and ∠OQT = 90°.
3. The angles of a quadrilateral add up to 360°: ∠PTQ + ∠POQ + 90° + 90° = 360°.
4. ∠PTQ = 360° − 180° − 110° = 70°.

Answer∠PTQ = 70° (option B).

Watch this explained “The four right angles”, 9:41 into Why the two from an outside point are always equally long

Open NCERT p. 151Checked by computer

1. PA and PB are tangents from the same external point P, so PA = PB, and OA = OB (radii). This makes triangles OAP and OBP congruent, so OP bisects ∠APB.
2. ∠APO = ∠BPO = 80° ÷ 2 = 40°.
3. OA is a radius to the point of contact A, so it is perpendicular to the tangent PA: triangle OAP is right-angled at A.
4. ∠POA = 90° − ∠APO = 90° − 40° = 50°.

Answer∠POA = 50° (option A).

Watch this explained “The corollary nobody asked for”, 6:28 into Why the two from an outside point are always equally long

Open NCERT p. 152One way to think about it

1. Let AB be a diameter of a circle with centre O, and let l and m be the tangents at A and at B.
2. The radius to a point of contact is perpendicular to the tangent there, so OA ⟂ l and OB ⟂ m.
3. A, O and B all lie on the one straight line AB, since AB is a diameter, so l and m are both perpendicular to this same line AB.
4. Two lines perpendicular to the same line are parallel to each other, so l is parallel to m.

In shortThe tangents at the two ends of a diameter are parallel, because both are perpendicular to the diameter itself.

Watch this explained “Exactly two, at the ends of one diameter”, 5:50 into A tangent as what a secant becomes when its two crossings merge

Open NCERT p. 152One way to think about it

1. Let l be a tangent to a circle with centre O, touching it at P, and let m be the line through P perpendicular to l.
2. The radius OP is already known to be perpendicular to the tangent l at P.
3. Through a point on a line, only one line can be drawn perpendicular to it.
4. Since both OP and m are perpendicular to l at the same point P, they must be the very same line.
5. So m is the line OP, which passes through the centre O.

In shortThe perpendicular to the tangent at the point of contact is exactly the line joining that point to the centre, so it passes through the centre.

Watch this explained “Exactly one tangent, and the normal”, 10:00 into Why the radius meets it at a right angle, argued from shortest distance

Open NCERT p. 152Matches NCERT’s answer

1. The radius to the point of contact is perpendicular to the tangent, so the triangle formed by the centre, the point of contact and A is right-angled at the point of contact.
2. The distance OA (5 cm) is the hypotenuse, and the tangent length (4 cm) is one leg.
3. By Pythagoras: 5² = r² + 4², so r² = 25 − 16 = 9, giving r = 3 cm.

Answerradius = 3 cm.

Watch this explained “The length without drawing anything”, 7:21 into None, one or two, depending on where the point sits

Open NCERT p. 152Matches NCERT’s answer

1. Let O be the common centre, and let AB be the chord of the larger circle that touches the smaller circle at M.
2. AB is a tangent to the smaller circle at M, so the radius OM is perpendicular to AB, and OM = 3 cm.
3. A perpendicular from the centre to a chord bisects the chord, so M is the midpoint of AB: AM = MB.
4. In triangle OMA, the angle at M is 90° and OA = 5 cm (a radius of the larger circle) is the hypotenuse. By Pythagoras: OA² = OM² + AM², so 5² = 3² + AM², AM² = 25 − 9 = 16 and AM = 4 cm.
5. AB = 2 × AM = 2 × 4 = 8 cm.

AnswerThe length of the chord is 8 cm.

Watch this explained “A tangent and a chord at once”, 7:23 into Why the two from an outside point are always equally long

Open NCERT p. 152One way to think about it

1. Let the circle touch side AB at P, side BC at Q, side CD at R and side DA at S, as shown in the figure.
2. From each corner, the two tangent segments drawn to the circle are equal in length: AP = AS, BP = BQ, CQ = CR, DR = DS.
3. Add sides AB and CD: AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS), replacing each piece by its equal partner.
4. Regroup the four pieces: (AS + DS) + (BQ + CQ) = AD + BC.

In shortAB + CD = AD + BC, because every side is made of two tangent pieces from its two corners, and those same four pieces regroup into AD + BC.

Watch this explained “The adding-up fact”, 12:39 into Why the two from an outside point are always equally long

Open NCERT p. 152One way to think about it

1. Join OA, OB, and also OP and OQ, where XY touches the circle at P and X′Y′ touches it at Q.
2. At point A, the two tangents XY and AB meet. The line from an external point to the centre always bisects the angle between the two tangents drawn from it, so OA bisects ∠PAB.
3. In the same way, OB bisects ∠QBA at point B.
4. XY and X′Y′ are parallel, and AB crosses both, so the two co-interior angles ∠PAB and ∠QBA add up to 180°.
5. Since OA and OB bisect these two angles, ∠OAB + ∠OBA is half of (∠PAB + ∠QBA), which is half of 180°, giving 90°.
6. In triangle OAB, the three angles add to 180°, so ∠AOB = 180° − 90° = 90°.

In short∠AOB = 90°.

Watch this explained “The corollary nobody asked for”, 6:28 into Why the two from an outside point are always equally long

Open NCERT p. 152One way to think about it

1. Let P be an external point, and let the two tangents from P touch the circle at Q and at R.
2. Join OQ and OR (radii) and OP, forming the quadrilateral OQPR.
3. The radius to a point of contact is perpendicular to the tangent there, so ∠OQP = 90° and ∠ORP = 90°.
4. The four angles of quadrilateral OQPR add up to 360°.
5. Subtracting the two right angles: ∠QPR + ∠QOR = 360° − 90° − 90° = 180°.

In short∠QPR (between the two tangents) and ∠QOR (at the centre, between the two radii to the points of contact) add up to 180° — they are supplementary.

Watch this explained “The four right angles”, 9:41 into Why the two from an outside point are always equally long

Open NCERT p. 152One way to think about it

1. Let ABCD be a parallelogram whose four sides all touch a circle.
2. As with any quadrilateral circumscribing a circle, the two tangent pieces from each corner are equal, so AB + CD = AD + BC (the same reasoning as in Q.8).
3. In a parallelogram, opposite sides are already equal: AB = CD and AD = BC.
4. Putting these together, AB + CD = AD + BC becomes AB + AB = AD + AD, that is, 2·AB = 2·AD, so AB = AD.
5. So all four sides are equal (AB = CD, AD = BC, and AB = AD), which makes ABCD a rhombus.

In shortA parallelogram circumscribing a circle must be a rhombus, because the tangent-length identity forces its two different side-lengths to be equal.

Watch this explained “The adding-up fact”, 12:39 into Why the two from an outside point are always equally long

Open NCERT p. 152Matches NCERT’s answer

1. Let the circle, with centre O, touch BC at D, CA at E and AB at F. Tangents from the same point are equal: BF = BD = 8 cm, CE = CD = 6 cm, and AF = AE = x cm (unknown).
2. So AB = x + 8, AC = x + 6 and BC = 8 + 6 = 14 cm.
3. Area, first way: join OA, OB and OC. The radius to each point of contact is perpendicular to that side, so each of the triangles OBC, OCA and OAB has height 4 cm. Area of ABC = ½ × 4 × 14 + ½ × 4 × (x + 6) + ½ × 4 × (x + 8) = 2 × (2x + 28) = 4(x + 14).
4. Area, second way, by Heron's formula: s = (AB + BC + CA) ÷ 2 = (2x + 28) ÷ 2 = x + 14, with s − BC = x, s − CA = 8 and s − AB = 6. Area = √[(x + 14) × x × 8 × 6] = √[48x(x + 14)].
5. Equate the two and square: 16(x + 14)² = 48x(x + 14). Divide both sides by 16(x + 14), which is not zero: x + 14 = 3x, so x = 7.
6. AB = 7 + 8 = 15 cm and AC = 7 + 6 = 13 cm. Check: sides 15, 14 and 13 give s = 21 and area √(21 × 6 × 7 × 8) = 84, and 84 ÷ 21 = 4, the given radius.

AnswerAB = 15 cm and AC = 13 cm.

Watch this explained “The triangle round a circle”, 13:56 into Why the two from an outside point are always equally long

Open NCERT p. 152One way to think about it

1. Let ABCD be a quadrilateral circumscribing a circle with centre O, and join OA, OB, OC and OD.
2. The line from any corner to O bisects the quadrilateral's angle at that corner (the same bisector fact used for two tangents from one point), so ∠OAB = A÷2, ∠OBA = B÷2, ∠OCD = C÷2, ∠ODC = D÷2, calling the quadrilateral's own angles A, B, C, D.
3. In triangle OAB, the angles add to 180°: ∠AOB = 180° − (A÷2 + B÷2).
4. In triangle OCD, likewise: ∠COD = 180° − (C÷2 + D÷2).
5. Adding these two: ∠AOB + ∠COD = 360° − (A + B + C + D)÷2.
6. The four angles of any quadrilateral add to 360°, so (A+B+C+D)÷2 = 180°.
7. So ∠AOB + ∠COD = 360° − 180° = 180°, and the same argument gives ∠BOC + ∠AOD = 180°.

In shortOpposite angles at the centre, ∠AOB with ∠COD, and ∠BOC with ∠AOD, are each supplementary (add up to 180°).

Watch this explained “The corollary nobody asked for”, 6:28 into Why the two from an outside point are always equally long