| Exercise 11.1 Exercise 11.2 |
Exercise 11.2
Question 1
From a point P, the length of the tangent to a circle is 15 cm, and the distance of P from the centre of the circle is 17cm. Then what is the radius of the circle?
Sol :Let the centre of circle be O so that PO = 17 cm
Tangent is PB whose length is 15 cm
OB is radius as shown
Now we know that radius is perpendicular to point of contact
Hence OB is perpendicular to PB
Hence ∠PBO = 90°
Consider ΔPBO
Using Pythagoras theorem
⇒ PB2 + OB2 = PO2
⇒ 152 + OB2 = 172
⇒ OB2 = 172 - 152
⇒ OB2 = 289 – 225
⇒ OB2 = 64
⇒ OB = ±8
As length cannot be negative
⇒ OB = 8 cm
Hence length of radius is 8 cm
Question 2
What is the distance between two parallel tangents of a circle of radius 10cm?
Sol :
O is the centre of circle and tangents from point A and B are parallel
We know that the line joining point of contacts of two parallel tangents (here AB) passes through the centre
And as a line from the centre is perpendicular to tangent, hence that line(AB) will be the distance between parallel tangents
AB passes through centre O hence AB is also the diameter of the circle
Hence the distance between the two parallel tangents will be the diameter of the circle
Radius is given 10 cm
Hence diameter of circle = 2 × radius
Hence AB = 2 × 10
⇒ AB = 20 cm
Hence distance between parallel tangents is 20 cm
Question 3
If the distance between two parallel tangents of a circle is 10cm, what is the radius of the circle?
Sol :
O is the centre of circle and tangents from point A and B are parallel
We know that the line joining point of contacts of two parallel tangents (here AB) passes through the centre
And as a line from the centre is perpendicular to tangent, hence that line(AB) will be the distance between parallel tangents
AB passes through centre O hence AB is also the diameter of the circle
Hence the distance between the two parallel tangents will be the diameter of the circle
The distance AB = 10 cm in diameter of the circle
Hence radius will be half of the diameter which is 5 cm
Question 4
The length of the tangent from a point A at a distance of 13cm from the centre of the circle is 12cm. What is the radius of the circle?
Sol :Let the centre of circle be O so that AO = 13 cm
Tangent is AB whose length is 12 cm
OB is radius as shown
Now we know that radius is perpendicular to point of contact
Hence OB is perpendicular to AB
Hence ∠ABO = 90°
Consider ΔABO
Using Pythagoras theorem
⇒ AB2 + OB2 = AO2
⇒ 122 + OB2 = 132
⇒ OB2 = 132 - 122
⇒ OB2 = 169 – 144
⇒ OB2 = 25
⇒ OB = ±5
As length cannot be negative
⇒ OB = 5 cm
Hence length of radius is 5 cm
Question 5 A
In the given figure if PA=20 cm, what is the perimeter of ΔPQR.
Sol :From P we have tangents PA and PB
Hence PA = PB …tangents from same point are equal …(a)
Point Q is on PA
From Q we have tangents QA and QC
⇒ QA = QC …tangents from same point are equal …(i)
Point R is on PB
From R we have two tangents RC and RB
⇒ RC = RB … tangents from same point are equal …(ii)
Consider ΔPQR
⇒ perimeter of ΔPQR = PQ + QR + PR
From figure QR = QC + CR
⇒ perimeter of ΔPQR = PQ + QC + CR + PR
Using (i) and (ii)
⇒ perimeter of ΔPQR = PQ + QA + RB + PR
From figure we have
PQ + QA = PA and RB + PR = PB
⇒ perimeter of ΔPQR = PA + PB
Using (a)
⇒ perimeter of ΔPQR = PA + PA
⇒ perimeter of ΔPQR = 2(PA)
PA is 20 cm given
⇒ perimeter of ΔPQR = 2 × 20
⇒ perimeter of ΔPQR = 40 cm
Question 5 B
In the given figure if ∠ATO=40°, find ∠AOB.
Sol :∠ATO = 40° …given
From T we have two tangents TA and TB
We know that if we join point T and centre of circle O then the line TO divides the angle between tangents
⇒ ∠ATO = ∠OTB = 40° …(i)
∠OAT = ∠OBT = 90° …radius is perpendicular to tangent …(ii)
Consider quadrilateral OATB
⇒ ∠OAT + ∠ATB + ∠TBO + ∠AOB = 360°…sum of angles of quadrilateral
From figure ∠ATB = ∠ATO + ∠OTB
⇒ ∠OAT + ∠ATO + ∠OTB + ∠TBO + ∠AOB = 360°
Using (i) and (ii)
⇒ 90° + 40° + 40° + 90° + ∠AOB = 360°
⇒ 260° + ∠AOB = 360°
⇒ ∠AOB = 100°
Question 6
In the figure PA and PB are tangents to the circle. If ∠APO=30°, find ∠AOB
Sol :∠APO = 30° …given
From P we have two tangents PA and PB
We know that if we join point P and centre of circle O then the line PO divides the angle between tangents
⇒ ∠APO = ∠OPB = 30° …(i)
∠OAP = ∠OBP = 90° …radius is perpendicular to tangent …(ii)
Consider quadrilateral OAPB
⇒ ∠OAP + ∠APB + ∠PBO + ∠AOB = 360°…sum of angles of quadrilateral
From figure ∠APB = ∠APO + ∠OPB
⇒ ∠OAP + ∠APO + ∠OPB + ∠PBO + ∠AOB = 360°
Using (i) and (ii)
⇒ 90° + 30° + 30° + 90° + ∠AOB = 360°
⇒ 240° + ∠AOB = 360°
⇒ ∠AOB = 120°
Hence ∠AOB is 120°
Question 7
AB and CD are two common tangents of two circles which touch each other at C. If D lies on AB and CD=5 cm, then what is the length of AB.
Sol :
DC and AB are tangents given to both circle
Point D is on AB which means DA and DB are also tangents to both circle
Now from point D, we have two tangents to bigger circle which are DA and DC
⇒ DA = DC …tangents from a point to a circle are equal
⇒ DA = 5 cm …DC is 5 cm given…(i)
Also from point D, we have two tangents to smaller circle which are DB and DC
⇒ DB = DC …tangents from a point to a circle are equal
⇒ DB = 5 cm …DC is 5 cm given…(ii)
Now as point D is on AB from the figure we can say that DA + DB = AB
⇒ AB = DA + DB
Using (i) and (ii)
⇒ AB = 5 + 5
⇒ AB = 10 cm
Hence the length of tangent AB is 10 cm
Question 8
In the given figure, ∠BPT=50°. What is the measure of ∠OPB?
Sol :Answer :
Sol :
PQ is tangent to circle and OP is radius
⇒ ∠OPQ= 90° …radius is perpendicular to tangent
⇒ ∠PQO = 30° …given
Consider ΔOPQ
⇒ ∠OPQ + ∠PQO + ∠POQ = 180° …sum of angles of triangle
⇒ 90° + 30° + ∠POQ = 180°
⇒ 120° + ∠POQ = 180°
⇒ ∠POQ = 60°
Hence ∠POQ is 60°
Consider ABCD as a parallelogram touching the circle at points P, Q, R and S as shown
As ABCD is a parallelogram opposites sides are equal
⇒ AB = CD …(a)
⇒ AD = BC …(b)
AP and AS are tangents from point A
⇒ AP = AS …tangents from point to a circle are equal…(i)
BP and BQ are tangents from point B
⇒ BP = BQ …tangents from point to a circle are equal…(ii)
CQ and CR are tangents from point C
⇒ CR = CQ …tangents from point to a circle are equal…(iii)
DR and DS are tangents from point D
⇒ DR = DS …tangents from point to a circle are equal…(iv)
Add equation (i) + (ii) + (iii) + (iv)
⇒ AP + BP + CR + DR = AS + DS + BQ + CQ
From figure AP + BP = AB, CR + DR = CD, AS + DS = AD and BQ + CQ = BC
⇒ AB + CD = AD + BC
Using (a) and (b)
⇒ AB + AB = AD + AD
⇒ 2AB = 2AD
⇒ AB = AD
AB and AD are adjacent sides of parallelogram which are equal hence parallelogram ABCD is a rhombus
Hence if all sides of a parallelogram touch a circle then that parallelogram is a rhombus
We can see that only 2 tangents can be drawn from an external point (here A) to a circle
Every other line either don intersects the circle or intersects at two points hence there can be only 2 tangents from external point to a circle
From P we have two tangents PA and PB
The tangents from an external point to a circle are equal
Hence PA = PB
PA = 6 cm given
Hence PB is also 6 cm
Only one tangent can be drawn from a point on the circle
Every other line will intersect the circle at two points which won't be a tangent
Touching three sides internally of a triangle which means the circle is inside the triangle hence it is called incircle of triangle
A circle is lying outside the triangle having one side of the triangle as tangent and also other two sides as tangent when they are extended.
There are three sides of a triangle hence there can be three excircles.
The tangents at the extremities of diameter are parallel to each other
Proof:
Consider a circle with centre O and diameter AB having tangents as PA and RB as shown
∠OAP = 90° …radius is perpendicular to the tangent at the point of contact
∠RBO = 90° … radius is perpendicular to the tangent at the point of contact
⇒ ∠OAP = ∠RBO
∠OAP and ∠RBO are alternate angles between lines PA and RB having transversal as AB
As alternate angles are equal lines, PA and RB are parallel
From P we have two tangents PM and PN
We know that if we join point P and centre of circle O then the line PO divides the angle between tangents
⇒ ∠MPO = ∠NPO …(a)
Consider ΔPNO
⇒ ∠PON = 50° …given
As radius ON is perpendicular to tangent PN
⇒ ∠PNO = 90°
Now
⇒ ∠PON + ∠PNO + ∠NPO = 180° …sum of angles of triangle
⇒ 50° + 90° + ∠NPO = 180°
⇒ 140° + ∠NPO = 180°
⇒ ∠NPO = 40° …(i)
From figure
⇒ ∠MPN = ∠MPO + ∠NPO
Using (a)
⇒ ∠MPN = ∠NPO + ∠NPO
⇒ ∠MPN = 2∠NPO
Using (i)
⇒ ∠MPN = 2 × 40°
⇒ ∠MPN = 80°
Hence ∠MPN is 80°
Sol :
a) we have to find ∠PTQ which is the angle between the tangents TP and TQ
∠OPT = ∠OQT = 90° …radius is perpendicular to tangent at point of contact
∠POQ = 90° …given
Consider quadrilateral POQT
⇒ ∠OPT + ∠OQT + ∠POQ + ∠PTQ = 360° …sum of angles of quadrilateral
⇒ 90° + 90° + 90° + ∠PTQ = 360°
⇒ 270° + ∠PTQ = 360°
⇒ ∠PTQ = 90°
Hence angle between tangents is 90°
Sol :
∠OPQ = 20° …given
Radius OP is perpendicular to tangent PA at point of contact A
⇒ ∠APO = 90°
From figure ∠APO = ∠APQ + ∠OPQ
⇒ 90° = ∠APQ + 20°
⇒ ∠APQ = 70° …(a)
Consider ΔAPQ
AP and AQ are tangents to circle from A
Tangents from a point to a circle are equal
⇒ AP = AQ
hence ΔAPQ is a isosceles triangle
⇒ ∠APQ = ∠AQP
Using (a)
⇒ ∠AQP = 70°
Now
⇒ ∠APQ + ∠AQP + ∠PAQ = 180° …sum of angles of triangle
⇒ 70° + 70° + ∠PAQ = 180°
⇒ 140° + ∠PAQ = 180°
⇒ ∠PAQ = 40°
Hence ∠PAQ is 40°
PT is tangent to circle and OP is radius
⇒ ∠OPT = 90° …radius is perpendicular to tangent
From figure
⇒ ∠OPT = ∠OPB + ∠BPT
⇒ 90° = ∠OPB + 50° …∠BPT is 50° given
⇒ ∠OPB = 40°
Hence ∠OPB is 40°
⇒ ∠OPT = 90° …radius is perpendicular to tangent
From figure
⇒ ∠OPT = ∠OPB + ∠BPT
⇒ 90° = ∠OPB + 50° …∠BPT is 50° given
⇒ ∠OPB = 40°
Hence ∠OPB is 40°
Question 9
In the given figure, a measure of ∠POQ is…..
Sol :
PQ is tangent to circle and OP is radius
⇒ ∠OPQ= 90° …radius is perpendicular to tangent
⇒ ∠PQO = 30° …given
Consider ΔOPQ
⇒ ∠OPQ + ∠PQO + ∠POQ = 180° …sum of angles of triangle
⇒ 90° + 30° + ∠POQ = 180°
⇒ 120° + ∠POQ = 180°
⇒ ∠POQ = 60°
Hence ∠POQ is 60°
Question 10
If all sides of a parallelogram touch a circle, then that parallelogram is….
Sol :Consider ABCD as a parallelogram touching the circle at points P, Q, R and S as shown
As ABCD is a parallelogram opposites sides are equal
⇒ AB = CD …(a)
⇒ AD = BC …(b)
AP and AS are tangents from point A
⇒ AP = AS …tangents from point to a circle are equal…(i)
BP and BQ are tangents from point B
⇒ BP = BQ …tangents from point to a circle are equal…(ii)
CQ and CR are tangents from point C
⇒ CR = CQ …tangents from point to a circle are equal…(iii)
DR and DS are tangents from point D
⇒ DR = DS …tangents from point to a circle are equal…(iv)
Add equation (i) + (ii) + (iii) + (iv)
⇒ AP + BP + CR + DR = AS + DS + BQ + CQ
From figure AP + BP = AB, CR + DR = CD, AS + DS = AD and BQ + CQ = BC
⇒ AB + CD = AD + BC
Using (a) and (b)
⇒ AB + AB = AD + AD
⇒ 2AB = 2AD
⇒ AB = AD
AB and AD are adjacent sides of parallelogram which are equal hence parallelogram ABCD is a rhombus
Hence if all sides of a parallelogram touch a circle then that parallelogram is a rhombus
Question 11
From an external point…… tangents can be drawn to a circle.
Sol :We can see that only 2 tangents can be drawn from an external point (here A) to a circle
Every other line either don intersects the circle or intersects at two points hence there can be only 2 tangents from external point to a circle
Question 12
From an external point P two tangents PA and PB have been drawn. IFPA = 6cm, then what is the length of PB.
Sol :From P we have two tangents PA and PB
The tangents from an external point to a circle are equal
Hence PA = PB
PA = 6 cm given
Hence PB is also 6 cm
Question 13
How many tangents can be drawn at a point on a circle ?
Sol :Only one tangent can be drawn from a point on the circle
Every other line will intersect the circle at two points which won't be a tangent
Question 14
What is the name of the circle touching the three sides of a triangle internally?
Sol :Touching three sides internally of a triangle which means the circle is inside the triangle hence it is called incircle of triangle
Question 15
How many excircles can be drawn to a triangle?
Sol :A circle is lying outside the triangle having one side of the triangle as tangent and also other two sides as tangent when they are extended.
There are three sides of a triangle hence there can be three excircles.
Question 16
What is the relation between the tangents at the extremities of a diameter of a circle?
Sol :The tangents at the extremities of diameter are parallel to each other
Proof:
Consider a circle with centre O and diameter AB having tangents as PA and RB as shown
∠OAP = 90° …radius is perpendicular to the tangent at the point of contact
∠RBO = 90° … radius is perpendicular to the tangent at the point of contact
⇒ ∠OAP = ∠RBO
∠OAP and ∠RBO are alternate angles between lines PA and RB having transversal as AB
As alternate angles are equal lines, PA and RB are parallel
Question 17
O is the centre of a circle. From an external point, P two tangents PM and PN have been drawn which touch the circle at M and N. If ∠PON = 50°, then find the value of ∠MPN.
Sol :From P we have two tangents PM and PN
We know that if we join point P and centre of circle O then the line PO divides the angle between tangents
⇒ ∠MPO = ∠NPO …(a)
Consider ΔPNO
⇒ ∠PON = 50° …given
As radius ON is perpendicular to tangent PN
⇒ ∠PNO = 90°
Now
⇒ ∠PON + ∠PNO + ∠NPO = 180° …sum of angles of triangle
⇒ 50° + 90° + ∠NPO = 180°
⇒ 140° + ∠NPO = 180°
⇒ ∠NPO = 40° …(i)
From figure
⇒ ∠MPN = ∠MPO + ∠NPO
Using (a)
⇒ ∠MPN = ∠NPO + ∠NPO
⇒ ∠MPN = 2∠NPO
Using (i)
⇒ ∠MPN = 2 × 40°
⇒ ∠MPN = 80°
Hence ∠MPN is 80°
Question 18 A
In the given figure, two radii OP and OQ of a circle are mutually perpendicular. What is the degree measure of the angle between tangents drawn to the circle at P and Q?
Sol :
a) we have to find ∠PTQ which is the angle between the tangents TP and TQ
∠OPT = ∠OQT = 90° …radius is perpendicular to tangent at point of contact
∠POQ = 90° …given
Consider quadrilateral POQT
⇒ ∠OPT + ∠OQT + ∠POQ + ∠PTQ = 360° …sum of angles of quadrilateral
⇒ 90° + 90° + 90° + ∠PTQ = 360°
⇒ 270° + ∠PTQ = 360°
⇒ ∠PTQ = 90°
Hence angle between tangents is 90°
Question 18 B
Centre of the circle is O and AP, and AQ is tangent of the circle. If ∠OPQ=20°, then what is the value of ∠PAQ?Sol :
∠OPQ = 20° …given
Radius OP is perpendicular to tangent PA at point of contact A
⇒ ∠APO = 90°
From figure ∠APO = ∠APQ + ∠OPQ
⇒ 90° = ∠APQ + 20°
⇒ ∠APQ = 70° …(a)
Consider ΔAPQ
AP and AQ are tangents to circle from A
Tangents from a point to a circle are equal
⇒ AP = AQ
hence ΔAPQ is a isosceles triangle
⇒ ∠APQ = ∠AQP
Using (a)
⇒ ∠AQP = 70°
Now
⇒ ∠APQ + ∠AQP + ∠PAQ = 180° …sum of angles of triangle
⇒ 70° + 70° + ∠PAQ = 180°
⇒ 140° + ∠PAQ = 180°
⇒ ∠PAQ = 40°
Hence ∠PAQ is 40°
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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