CHAPTER 10: CIRCLE
TERMS RELATED TO CIRCLE
Definition: Suppose a point is fixed (like a pole) and another point (like a goat) She is moving around the pole, tied with a rope that is fully stretched. If the goat moves one round and we trace the path of the goat, then this path is known as a circle. The length of the rope from the pole is called the radius of the circle.
Chord: A line joining the two points on the circle's boundary is known as a chord; it does not cross the circle.
Secant: A line that crosses two points on a circle is known as a secant.
EXERCISE 10.1
Q1: How many tangents can a circle have?
Ans: A tangent is a line outside the circle that touches the circle at a single point.
On a circle, we can draw many tangents. Here in the above figure, tangents are touching the circle at A, E, F, and D.
Similarly, we can draw many tangents taking infinite points.
The answer is infinite tangents.
Question 2 Fill in the blanks:
(i) A tangent to a circle intersects it in ……one…… point(s).(ii) A line intersecting a circle in two points is called a …secant….
(iii) A circle can have ……two…………. parallel tangents at the most.(iv) The common point of a tangent to a circle and the circle is called. Point of contact.
Question 3: A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 12 cm. Length PQ is
From the given information, OQ is the hypotenuse, using the Pythagorean theorem
PQ is solved in the figure.
Question 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.
Theorem: A tangent drawn to a circle is perpendicular to the radius.
To prove: OA is perpendicular to AB
Proof: In the triangle OAB
OB > OA
As B lies outside the circle and OA is the radius
hence, OB > OA
Similarly, we can consider other points on AB that are greater than OA (radius).
Therefore, if we compare OA with the distances of the other points lying on AB, OA is the shortest distance.
We know that the shortest distance of a point from a straight line is perpendicular to the line.
We know that the shortest distance of a point from a straight line is perpendicular to the line.
So, OA is perpendicular to AB.
Important questions from the example for the board
Example: Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Given: AB is the chord of the larger circle, touching the inner circle at point P. OP is the radius of both circles.
To prove: AB is bisected at P or AP = PB
Construction: Join OA and OB
Proof: In ▲AOP and ▲BOP
OA = OB (radii)
OP = OP (common)
Angle OPA = Angle OPB (as the radius is perpendicular to the tangent, so each angle is 90 degrees).
▲AOP ≋▲BOP (RHS)
Therefore, AP = PB (c.p.c.t.)
Hence, proved
EXERCISE 10.2
1. From a point Q, the length of the tangent to a circle is 24 cm, and the distance of Q from the center is 25 cm. The radius of the circle is
Solution
Option A is correct.
Q2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ã„¥ POQ = 110°, then PTQ is equal to
Solution
ã„¥OPT =ã„¥OQP = 90
ã„¥POQ = 110
ã„¥PTQ = 360 - (90 + 90 + 110)
= 70
Option B is correct.
Question 3: If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ã„¥POA is equal to
In Quadrilateral AOBP, AOBPAngle AOB = 360 - (90 + 90 + 80).
= 100
Angle POA = Angle AOB/2 = 100/2 = 50
Option A is correct.
Question 4: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Given: Two tangents AB and CD, touching the circle at points P and Q
To prove: AB is parallel to CD
Proof: In the figure, OP and OQ are the radii, and OP and OQ are perpendicular to the tangent at points P and Q
Angle P + Angle Q = 90 + 90
= 180
As, PQ is a diameter also, it can be considered as a transversal for the AB and CD , which are intersected at points P and Q by PQ
Some of the co-interior angles OQB and OPB = 180
Hence, AB and CD are parallel.
Question 5: Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the center.
.
Given: PQ is a tangent and touching a circle at point P.
To prove: PO passes through the center of the circle.
Proof: This question can be solved by the method of contradiction.
If possible, let O not be the center of the circle; let O' be another point in the circle that is the center.
Given that OP is perpendicular to PQ
Angle OPQ = 90
And if O' is the center of the circle, then O'P will be perpendicular to P.
Angle O'PQ = 90 (radius O'P will be perpendicular to tangent)
In this case
Angle OPQ = 90 = Angle O'PQ
This can be possible only when O and O' are the same point or coincide with each other.
So, OP = O'P
Hence, O' and O are the same point
So, a perpendicular at the point of contact to a tangent passes through the center as a radius passes through the center.
Question6: . The length of a tangent from a point A at a distance 5 cm from the centre of the circle is 4
cm. Find the radius of the circle.
OA = 5 cm
AB = 4 cm
using the Pythagorean theorem
Question 7: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle that touches the smaller circle.
Solution 6:
OP = 3 cm, radius of the smaller circle
In triangle OAP
AO2 = AP2 + OP2
AP2 = 52- 42
= 32
AP = 3cm
As the radius divides the chord into two equal parts
AB = 2AP = 2 * 3 = 6 cm
Question 8: A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC
To prove: AB + CD = AD + BC
Proof:
Tangents drawn from the external point to a circle are equal in length, hence,
At point D
DR = DS
At point A
At point A
AS = AP
At point B
BP = BQ
At point C
CQ =CR
Now, taking the LHS and breaking into segments—
AB + CD = AP + PB + DR + RC
= AS + BQ + SD + CQ using the above conditions
= (AS + SD) + (BQ + CQ)
= AD + BC
Proved.
Question 9: In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with center O, and
another tangent AB with point of contact C intersects XY at A and X'Y' at B. Prove
that ∠AOB = 90°.
Given: XY is parallel to X'Y', which are parallel tangents. Another tangent, AB, touches the circle at C.
To prove: ∠AOB = 90°.
Solution:
XY and X'Y' are parallel tangents, and AB can be considered as a transversal
So,
∠PAC + ∠CBA = 180 degrees [1]
∠PAO = ∠CAO (line joining the centre and the external point bisects the angle between the tangents from the external points) [2]
Similarly,
∠OBQ = ∠OBC (reason as above in bracket) [3]
From [1], [2], and [3]
Divide [1] by 2
1/2∠PAC + 1/2∠CBA ∠CBA = 1/2 * 180 degrees
∠OAC + ∠OBC = 90 degrees
Now, in triangle OAB
∠AOB = 180 - (∠OAC + ∠OBC)
= 180 - 90 = 90
Hence, proved.
Question 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.
Given that: PA and PB are two tangents drawn from an external point P.
To prove: ∠APB + ∠AOB = 180 (supplementary)
Proof:
Since radii are perpendicular to the tangent
∠OAP = ∠OBP = 90
Therefore, in quadrilateral PAOB
∠APB + ∠PBO + ∠BOA + ∠OAP = 360 (sum of interior angles of a quadrilateral)
∠APB + ∠AOB + 90 + 90 = 360
∠APB + ∠AOB = 180
Hence, proved
Question 11:Prove that the parallelogram circumscribing a
circle is a rhombus.
To prove: ABCD is a rhombus.
Proof: Since the tangents drawn from the external point to a circle are equal in length,
Hence, from point A
AS = AP
From point D
DS = DR
From point C
CR= CQ
From point B
BP = BQ
In a parallelogram, opposite sides are equal.
In a parallelogram, opposite sides are equal.
AB + DC = (AP + PB) + (DR + RC)
= (AS + BQ) + (DS + CQ)
= (AS+DS) + (BQ+CQ)
= AD + BC
Or, we can write
2AB = 2AD (as AB = DC and AD = BC)
AB = AD
Similarly, we can write
DC = BC
Hence, from above, we can see that
AB = BC = CD = DA
All the sides are equal; hence, it is a rhombus.
Question 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. 10.14). Find the sides AB
and AC.
CD = 6 cm
BD = 8 cm
To find the AB and AC
Let AE = AF = x cm
as the tangents from the external point of the circle are equal in length [1]
AT C
CD = CE by [1]
BD = BF [1]
AC = CE + AE = 6 +x
AB = AF + FB = 8 + x
Perimeter of the triangle = AC + BC + AB
= (8+6) +(6+x) +(8+x)
= 28 + 2x
Semiperimeter (s) = (28+2x)/2 = 14 + x
Using Heron's formula, we find the area of the triangle
s = 14 + x
s-a = 14+x -(14) = x
s-b = 14+x - (6+x) = 8
s-c = 14 + x - (8 + x) = 6
Area of the triangle formula
Area = {s(s-a)(s-b)(s-c)}1/2
= {(14+x)*x *8*6}1/2
= {48x (14+x)}1/2
Again, the area of
triangle ABC = the Area of triangle OAB + the Area of triangle OBC the area of triangle AOC
= ½ (CB)(OD)
+ ½ (AB)(OF) +1/2(AC)(OE)
Since, OD = OF = OE = radii
= ½(OD)(AB+AC+CB)
= ½
(OD) (28+2x)
=
OD (14+x)
Comparing both the
area
OD (14+x) = {48x (14+x)}1/2
Given OD = 4cm
Squaring both sides
42
(14+x)2 = 48x (14+x)
ð
16(14+x){
14+x – 3x} = 0
ð
16(14+x)
(14-2x) = 0
ð
(14-2x)
= 0 gives x = 14/2 = 7
ð
And
14+x = 0 gives x= -14 ( rejected)
Hence,
x= 7 cm
AC=
AE +CE = 6+7 = 13 cm
AB = AF +FB = 8+7 = 15 cm
Ans. AB = 15 cm, AC = 13 cm
Question 13: Prove that opposite sides of a quadrilateral
circumscribing a circle subtend supplementary
angles at the centre of the circle. ( Important for the board exam)
Given: ABCD is a quadrilateral that circumscribes a circle with center O.
To prove: ∠ AOB +∠DOC = 180
and ∠AOD + ∠BOC = 180
Construction: Join OA, OB, OC, and OD
Proof: As the tangents drawn from the external point to a circle are equally inclined to each other from the center
∠1 = ∠2
∠3 = ∠4
∠5 =∠6
∠7 =∠8
So, ∠1 +∠2 +∠3 +∠4 +∠5 + ∠6 +∠7 + ∠8 = 360 ( sum of angles around a point =360)
=> 2∠1 +2∠8 +2∠4 +2∠5 = 360
2(∠1 +∠8 +∠4 +∠5) = 360
(∠1 +∠8 +∠4 +∠5) = 360/2
(∠1 +∠8 +∠4 +∠5) = 180
∠ AOB +∠DOC = 180
Similarly second part can be shown.
--------------------------------------------Other questions-----------------------
Find angle POQ
Solution:
∠TPO = 90 ( angle between Tangent TP and radius OP)
∠ QPO = 90-50 = 40 degrees
OP = OQ
∠ OPQ = ∠ OQP = 40 degrees ( isosceles triangle OPQ)
So, ∠ POQ = 180 - (40 +40) = 100 degrees ( Angle sum property of triangle POQ)
Ans 40 degrees.
Q2: In the following figure, AP and AQ are the tangents to the circle. AB = 5cm, AC = 6cm, BC = 4cm, then the length of AP?
AB = 5cm
At point B
BP = BC = 4cm ( tangents from an external point to a circle are equal in length)
AP = AB +BP = 5 +4 = 9cm
Q3: Two circles of radii a and b, where a>b. The length of a chord of the larger circle that touches the other circle is -?
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