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Chapter 10 – Circles

Circle

NCERT TEXTBOOK QUESTIONS SOLVED
EXERCISE 10.1
Q.1.   How many tangents can a circle have?
Sol. A circle can have an infinite number of tangents.
Q.2.   Fill in the blanks:
        (i) A tangent to a circle intersects it in ……….. point(s).
        (ii) A line intersecting a circle in two points is called a……….. .
        (iii) A circle can have………..parallel tangents at the most.
        (iv) The common point of a tangent to a circle and the circle is called ……….. .
Sol. (i) exactly one        (ii) secant        (iii) two        (iv) point of contact.
Q.3.   A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length of PQ is:
        (A) 12 cm
(B) 13 cm
(C) 8.5
cm (D)
Q.4.   Draw a circle and two lines parallel to a given line such that one is a tangent and the other a secant to the circle.
Sol. We have the required figure.
        Here, l is the given line and a circle with centre 0 is drawn.
        The line PT is drawn which is parallel to l and tangent to the circle.
        Also, AB is drawn parallel to line l and is a secant to the circle.
EXERCISE 10.2
Q.1.   Choose the correct option:
        From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
        (A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Sol. QT is a tangent to the circle at T and OT is
        radius
        Also, OQ = 25 cm and QT = 24 cm
        ? Using Pythagoras theorem, we get
              OQ2 = QT2 + OT2
        ? OT2 = OQ2 – QT2
              = 252 – 242 = (25 – 24) (25 + 24)
              = 1 × 49 = 49 = 72
        ? OT = 7
        Thus, the required radius is 7 cm.
        ? The correct option is (A).
Q.2.   Choose the correct option:
        In figure, if TP and TQ are the two tangents to a circle with centre O so that LPOQ = 110°, then LPTQ is equal to
        (A) 60°
(B) 70°
(C) 80°
(D) 90°
Sol. ?TQ and TP are tangents to a circle with centre O.
        such that ?POQ = 110°
        ? OP ? PT and OQ ? QT
        ? ?OPT = 90° and ?OQT = 90°
        Now, in the quadrilateral TPOQ, we get
        ? ?PTQ + 90° + 110° + 90° = 360°
        ? ?PTQ + 290° = 360°
        ? ?PTQ = 360° – 290° = 70°
        Thus, the correct option is (B).
Q.3.   Choose the correct option:
        If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ?POA is equal to
        (A) 50°
(B) 60°
(C) 70°
(D) 80°
Sol. Since, O is the centre of the circle and two tangents from P to the circle are PA and PB.
        ? OA ? AP and OB ? BP
        ? ?OAP = ?OBP = 90°
        Now, in quadrilateral PAOB, we have:
        ?APB + ?PAO + ?AOB + ?PBO = 360°
        ? 80° + 90° + ?AOB + 90° = 360°
        ? 260° + ?AOB = 360°
        ? ?AOB = 360° – 260°
        ? ?AOB = 100°
        In rt ?OAP and rt ?OBP, we have
              OP = OP
              ?OAF = ?OBP
              OA = OB
        ? ? OAP ? ? OBP
        ? Their corresponding parts are equal
        ? ?POA = ?POB
        
        Thus, the option (A) is correct.
Q.4.   Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Sol. In the figure, we have:
        PQ is diameter of the given circle and O is its centre.
        Let tangents AB and CD be drawn at the end points of the diameter PQ.
        Since the tangent at a point to a circle is perpendicular to the radius through the point.
PQ ? AB ? ?APQ = 90°
        And PQ ? CD ? ?PQD = 90°
        ? ?APQ = ?PQD
        But they form a pair of alternate angles.
               AB ?? CD.
Q.5.   Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Sol. In the figure, the centre of the circle is 0 and tangent AR touches the circle at P. If possible, let PQ be perpendicular to AB such that it is not passing through O.
        Join OP.
        Since tangent at a point to a circle is perpendicular to the radius through that point,
        ? AB ? OP                i.e. ?OPB = 90°                …(1)
        But by construction,
              AB ? PQ ?                ?QPB = 90°                …(2)
        From (1) and (2),
              ?QPB = ?OPB
        which is possible only when O and Q coincide.
        Thus, the perpendicular at the point of contact to the tangent passes through the centre.
Q.6.   The length of a tangent from a point A at distance. 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Sol. ?The tangent to a circle is perpendicular to the radius through the point of contact.
              ?OTA = 90°
        Now, in the right ?OTA, we have:
              OP2 = OT2 + PT2
        ? 52 = OT2 + 42
        ? OT2 = 52 – 42
        ? OT2 = (5 – 4) (5 + 4)
        ? OT2 = 1 × 9 = 9 = 32
        ? OT = 3
        Thus, the radius of the circle is 3 cm.
Q.7.   Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Sol. In the figure, O is the common centre, of the given concentric circles.
        AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.
        Since OP is the radius of the smaller circle through P, the point of contact,
        ? OP ? AB
        ? ?APB = 90°
        Also, a radius perpendicular to a chord bisects the chord.
        
        Now, in right ?APO,
              OA2 = AP2 – OP2
        ? 52 = AP2 – 32
        ? AP2 = 52 – 32
        ? AP2 = (5 – 3) (5 + 3) = 2 × 8
        ? AP2 = 16 = (4)2
        ? AP = 4 cm
        
        Hence, the required length of the chord AB is 8 cm.
Q.8.   A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that:
              AB + CD = AD + BC
Sol. Since the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touch the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.
              AP = AS
              BP = BQ
              DR = DS
              CR = CQ
        Adding them, we get
        (AP + BP) + (CR + RD) = (BQ + QC) + (DS + SA)
        ? AB + CD = BC + DA
        which was to be proved.
Q.9.   In the figure, XY and X’Y’are two parallel tangents to a circle with centre 0 and another tangent AB with point of contact C intersecting XY at A and XY’ at B. Prove that ZAOB = W.
Sol. ?The tangents drawn to a circle from an external point are equal.
        ? AP = AC
        In ? PAO and ? AOC, we have:
              AO = AO                                          [Common]
              OP = OC                                          [Radii of the same circle]
              AP = AC
        ? ?PAO ? ?AOC
        ? ?PAO = ?CAO
              ?PAC = 2 ?CAO                                                        …(1)
        Similarly ?CBQ = 2 ?CBO                                          …(2)
        Again, we know that sum of internal angles on the same side of a transversal is 180°.
        ? ?PAC + ?CBQ = 180°
        ? 2?CAO + 2 ?CBO = 180°                                          [From (1) and (2)]
        
        ? 90° + ?AOB = 180°
        ? ?AOB = 180° – 90°
        ? LAOB = 90°.
Q.10.   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-segment joining the points of contact at the centre.
Sol. Here, let PA and PB be two tangents drawn from an external point P to a circle with centre O.
        Now, in right ? OAP and right ? OBP, we have
               PA = PB                                             [Tangents to circle from an external point P]
               OA = OB                                             [Radii of the same circle]
               OP = OP                                             [Comm]
        ? By SSS congruency,
               ? OAP ? OBP
        ? Their corresponding parts are equal.
               ?OAA = ?OPB
        And ?AOP = ?BOP
        ? ?APB = 2 ?OPA and ?AOS = 2 ?AOP
        But ?AOP = 90° – LOPA
        ? 2 ?AOP = 180° – 2 ?OPA
        ? ?AOB = 180° – ?APB
        ? ?AOB + ?APB = 180°.
Q.11.   Prove that the parallelogram circumscribing a circle is a rhombus. (CBSE 2012, CBSE Delhi 2014)
Sol. We have ABCD, a parallelogram which circumscribes a
        circle (i.e., its sides touch the circle) with centre D.
        Since tangents to a circle from an external point are equal in length
        ? AP = AS
               BP = BQ
               CR = CQ
               DR = DS
        Adding, we get
        (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
        ? AB + CD = AD + BC
        But AB = CD                                                        [opposite sides of ABCD]
        and BC = AD
        ? AB + CD = AD + BC ? 2 AB = 2 BC
        ? AB = BC
        Similarly AB = DA and DA = CD
        Thus, AB = BC = CD = AD
        Hence ABCD is a rhombus.
Q.12.   A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find tlic sides AR and AC.
Sol. Here ? ABC subscribe the circle with centre O.
        Also, radius = 4 cm
        ? The sides BC, CA and AB touch the circle at D, E and F respectively.
        ? BF = BD = 8 cm
              CE = CD = 6 cm
              AF = AE = x cm (say)
        ? The sides of the triangle are:
              14 cm, (x + 6) cm and (x + 8) cm
        Perimeter of ? ABC
              = [14 + (x + 6) + (x + 8)] cm
              = [14 + 6 + 8 + 2x] cm
              = 28 + 2x cm
        ? Semi perimeter of ? ABC
        
        ? S – AB = (14 + x) – (8 + x) = 6
              S – BC = (14 + x) – (14) = x
              S – AC = (14 + x) – (16 + x) = 8
        
        
        ? ar (?ABC) = ar (? OBC) + ar (? OCA) + ar (? OAB)
              = 28 cm2 + (2x + 12) cm2 + (2x + 16) cm2
              = (28 + 12 + 16) + 4x cm2
              = (56 + 4x) cm2
        From (1) and (2), we have:
        
        Squaring both sides
              (14 + x)2 = (14 + x) 3x
        ? 196 + x2 + 28x = 45x + 3x2
        ? 2x2 + 14x – 196 = 0 ? x2 + 7x – 98 = 0
        ? (x – 7) (x + 14) = 0
        ? Either x – 7 = 0 ? x = 7
        or x +14 = 0 ? x = (–14)
        But x = (– 14) is not required
        ? x = 7 cm
        Thus, AB = 8 + 7 = 15 cm
              BC = 8 + 6 = 14 cm
              CA = 6 + 7 = 13 cm.
Q.13.   Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Sol. We have a circle with centre O.
        A quadrilateral ABCD is such that the sides AB, BC, CD and DA touch the circle at P, Q,,F R and S respectively.
        Let us join OP, OQ, OR and OS. We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.
        ? ?1 = ?2
              ?3 = ?4
              ?5 = ?6 and ?7 = ?8
        Also, the sum of all the angles around a point is 360°.
        ? ?1 + ?2 + ?3 + ?4 + ?5 + ?6 + ?7 + ?8 = 360°
        ? 2 [?1 + ?8 + ?5 + ?4] = 360°
        ? (?1 + ?8 + ?5 + ?4) = 180° (1)
        And 2 [?2 + ?3 + ?6 + ?7] = 360°
        ? (?2 + ?3) + (?6 + ?7) = 180° (2)
        Since, ?2 + ?3 = ?AOB
               ?6 + ?7 = ?COD
               ?1 + ?8 = ?AOD
               ?4 + ?5 = ?BOC
        ? From (1) and (2), we have:
               ?AOD + ?BOC = 180° and
               ?AOB + ?COD = 180°
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