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CLASS 10 Circles Solutions MATHEMATICS
Chapter 10 – CIRCLES
In Q.1 to 3, choose the correct option and give justification.
1. 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
Answer: Option (A)
If OP is radius which is perpendicular to the tangent PQ. Q is the external point which is joined to the centre O.
So OP PQ
From the figure,
We have got △OPQ is a right angled triangle.
It is given that
OQ = 25 cm
and PQ = 24 cm
By using Pythagoras theorem in △OPQ,
OQ2 = OP2 + PQ2
(25)2 = OP2+ (24)2
625 = OP² + 576
OP2 = 625 – 576
OP2 = 49
OP = 7 cm
option A (7 cm) is the radius of the circle.
2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
Answer: Option (B) 70
From the question,
OP and OQ are radius of the circle which are perpendicular to tangents PT and TQ.
OP ⊥ PT and TQ ⊥ OQ
So, In Quadrilateral OPTQ
We know that the sum of the interior angles is 360.
∠OPT = ∠OQT = 90°
∠PTQ+∠POQ+∠OPT+∠OQT = 360°
putting the respective values we get,
∠PTQ +90°+110°+90° = 360°
∠PTQ = 70°
∠PTQ is 70° which is option B (70°).
3. 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
Answer: Option A (50°).
According to the question, OA and OB are radius of the circle which are perpendicular to tangents PB and PA.
So OA ⊥ PA and OB ⊥ PB
∠OBP = ∠OAP = 90°
in the quadrilateral AOBP,
The sum of all the interior angles will be 360°
∠AOB+∠OAP+∠OBP+∠APB = 360°
Putting their values,
∠AOB + 90 + 90 + 80 = 360
∠AOB + 260° = 360°
∠AOB = 100°
The triangles △OPB and △OPA.
AP = BP (Since the tangents from a point are always
OA = OB (Which are the radii of the circle)
OP = OP (It is the common side)
Now, we can say that triangles OPB and OPA are
Similar using SSS congruency.
∴△OPB ≅ △OPA
So, ∠POB = ∠POA
∠AOB = ∠POA + ∠POB
2 (∠POA) = ∠AOB
Putting the respective values,
- ∠POA = 100°/2
- = 50°
As ∠POA is 50°
Therefore, answer is Option A (50°).
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Construction – Draw a circle centered O with diameter PQ and draws two tangents AB and CD at points P and Q respectively.
To Prove – AB || CD
Now, both radii i.e. OP and OQ are perpendicular to
the tangents AB and CD respectively.
∠OPA = ∠OQC = ∠OPB = ∠OQD = 90°
∠APO = ∠DQO
and ∠BPO = ∠OQC (Since they are also alternate interior angles)
We can say that line PQ and RS are parallel to each other.
PQ || RS. (Hence Proved).
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.
Construction – Draw a circle with center O and a tangent AB which passes through L. Join O and L. Which will make radius OL.
Also let another center which is at point P. Join P to L. As PL also a radius.
To prove : O and P are coincident each other.
Since tangent at a point to a circle is perpendicular to the radius through the point.
OL AB So ∠OLB = 90 ͦ ………(1)
But if PL is also radius of circle then
∠PLB = 90 ͦ (by construction) ………..(2)
So by equation (1) and (2)
∠OLB = ∠PLB
But this is not possible that from same tangent point there is two perpendiculars.
Means O and P point are same points or we can say they coinside each other.
So we can say that –
the perpendicular at the point of contact to the tangent to a circle passes through the center.
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.
According to the diagram PA is the tangent of circle centered O, Join OA. And OP is the radius.
So, OP ⊥ PA
We know, OA (hypotenuse) = 5cm and
PA (perpendicular) = 4 cm
Now, △PAO is a right triangle,
OA2 = PA2 + OP2(Using Pythagoras theorem)
52 = 42 + OP2
25 = 16 + OP2
OP2 = 25-16
OP2 = 9
OP = 3 cm
So, the radius of the given circle i.e. OP is 3 cm.
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.
Draw two concentric circles with center O.
And also, draw a chord AB in the larger circle which
touches the smaller circle at a point L as shown
in the figure below.
According to the figure,
AB is tangent to the smaller circle to point L.
∴ OL ⊥ AB
Using Pythagoras theorem in triangle OPA,
52 = AL2+32
AL2 = 25-9
AL = 4 cm
Now, AL = LB ( OL ⊥ AB, and The perpendicular from the center of the circle bisects the chord)
So, AB = 2AL
= 8 cm
So, the length of the chord of the larger circle
is 8 cm.
8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC
By the figure, we can get
DR = DS …………(1) the lengths of tangents drawn from an external point to a circle are equal.
BP = BQ …………(2)
AP = AS …………(3)
CR = CQ …………(4)
adding LHS and RHS of equation (1),(2),(3),(4)
DR+BP+AP+CR = DS+BQ+AS+CQ
(DR+CR) + (BP+AP) = (CQ+BQ) + (DS+AS)
9. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.
In the above figure, we will join OC
Then we will get-
Now, In triangles △OPA and △OCA
OP = OC (Radii of the same circle)
AO = AO (Common side)
AP = AC (length of tangents are equal from point A)
So, by sss congruency
△OPA ≅ △OCA
△OQB ≅ △OCB
∠POA = ∠COA … (Eq i)
∠QOB = ∠COB … (Eq ii)
the line POQ is a straight line,
it can be considered as a diameter of the circle.
So, ∠POA +∠COA +∠COB +∠QOB = 180°
Now, from equations (i) and (ii) we get,
2∠COA + 2∠COB = 180°
∠COA + ∠COB = 90°
∴∠AOB = 90°
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 center.
draw a circle with centre O. From an external point B draw two tangents BC and BA at point A and C where OC and OB are radius of circle.
Now, The diagram is as follows:
We get quadrilateral OABC
In which ∠ OCB = 90 and ∠OAB = 90
We know that the tangent at any point of a circle is perpendicular to radius through the point of contact.
in the quadrilateral OAPB,
∴∠OCB+∠COA +∠OAB +∠ABC = 360°
(Since the sum of all interior angles of quadrilateral is 360°)
By putting the values we get,
90 + ∠COA + 90 + ∠ABC = 360°
180 + ∠COA + ∠ABC = 360°
∠COA + ∠ABC = 180° (Hence proved).
11. Prove that the parallelogram circumscribing a circle is a rhombus.
As in the figure let a parallelogram ABCD which is circumscribing a circle with a center O.
Now, ABCD is a parallelogram,
So AB = CD and BC = AD.
And we know that,
DS = DP …… (1) The lengths of tangents drawn from an external point to a circle are equal.
AS = AR …… (2)
BQ = BR …… (3)
CQ = CP …… (4)
Adding equations (1), (2), (3), (4) we get,
DS + AS + BQ + CQ = DP + AR + BR + CP
By rearranging, we get,
(DS + AS) + (BQ + CQ) = (DP + CP) + (AR + BR)
Again by rearranging them we get,
AD + BC = CD + AB
AB = CD and BC = AD,
the above equation becomes
2AD = 2CD
∴ AD = CD
AB = BC = CD = DA,
it can be said that ABCD is a rhombus.
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 Fig.). Find the sides AB and AC.
From the figure:
ABC is given, And a circle is inscribed in the triangle.
CD = 6 cm , BD = 8 cm.
CF = CD = 6 cm (The lengths of tangents drawn from an external point to a circle are equal.)
BE = BD = 8 cm
Now from the above figure let AF be x cm , So –
AE = AF = x
We can say that,
AB = EB+AE
CA = CF+FA
BC = DC+BD
= 6+8 = 14
We will calculate now are of triangles by herons formula
s = (AB+CA+BC)/2 (Where s = semi perimeter of triangle)
By putting the values we get,
s = (x + 8 + x + 6 + 14)/2
s = (2x + 28)/ 2
s = x + 14
Where a = BC, b = AC, c = AB
By putting the values,
= ……… (i)
the area of △ABC
= area of (△AOC + △BOC + △COA)
= [(½×(x+8)×4)+ (½×(x+6)×4) + ([(½×(14)×4)
= [ 2x + 16 + 2x + 12 + 28 ]
= 56+4x ………….. (ii)
Now from (i) and (ii) we get,
= = 56+4x
squaring both the sides,
48x(14+x) = (56+4x)2
48x = [4(14+x)]2/(14+x)
48x = 16(14+x)
48x = 224+16x
32x = 224
x = 7 cm
AB = 8+x
= 8 + 7 = 15 cm
And, CA = x+6
= 7+6 = 13 cm,
So answer is AB = 15 cm and CA = 13 cm.
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution: To prove –
Construct a quadrilateral ABCD and circumscribed a circle with centre O which touches the circle at point P, Q, R, and S. Now, after joining the vertices of ABCD
We will get the figure:
Lets take the triangles OAP and OAS,
AP = AS (The lengths of tangents drawn from an external point to a circle are equal.)
OA = OA (common side)
OP = OS (radii of the circle)
So, by SSS congruency △OAP ≅ △OAS
So, ∠POA = ∠AOS
Means ∠2 = ∠1
∠3 = ∠4
∠5 = ∠6
∠8 = ∠7
by adding all these angles we get,
∠1+∠2+∠3 +∠4 +∠5+∠6+∠7+∠8 = 360°
Now by rearranging,
(∠2+∠2)+(∠3+∠3)+(∠6+∠6)+(∠7+∠7) = 360°
2∠2+2∠3+2∠6+2∠7 = 360°
2(∠2+∠3+∠6+∠7) = 360°
∠2+∠3+∠6+∠7 = 180°
(∠2+∠3)+(∠6+∠7) = 180°
∠AOB+∠COD = 180°
We can prove that ∠BOC+∠AOD = 180°
The opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the center of the circle.