NCERT TEXTBOOK QUESTIONS SOLVED
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
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.
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
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
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
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.
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°