Circles, Class 10, Maths Detailed Chapter Notes PDF Download

INTRODUCTION

In class IX, we have studied that a circle is a collection of all points in a plane which are at a constant distance from a fixed point. The fixed point is called the centre and the constant distance is known as the radius. We have also studied various terms related to a circle like chord, segment, sector, arc etc. Now, we shall study properties of a line touching a circle at one point.

RECALL

Circle

A circle is the locus of a point which moves in such a way that it is always at the constant distance from a fixed point in the plane. The fixed point 'O' is called the centre of the circle. The constant distance 'OA' between the centre (O) and the moving point (A) is called the Radius of the circle.

Circumference

The distance round the circle is called the circumference of the circle.

2πr = circumference of the circle

= Perimeter of the circle.
= boundary of the circle
r is the radius of the circle.

Chord

The chord of a circle is a line segment joining any two points on the circumference. AB is the chord of the circle with centre O. In fig. AB is the chord of the circle.

Diameter

A line segment passing through the centre of the circle and havingits end points on the circle is called diameter. If r is the radius of the circle then the diameter of the circle is twice the radius i.e., d = 2r

AOB is a diameter of the circle whose centre is O

AOB = OA + OB = r + r = 2r.

Arc of a circle

If P and Q be any two points on the circle then the circle is divided into two pieces each of which is an arc. Now we denote the arc from P to Q in counter clock-wise direction by PQ and the arc from Q to P in clock-wise direction by QP.

Sector of a circle

The part of a circle bounded by two radii and arc is called sector.In fig, the part of the plane region enclosed by AB and its bounding radii OA and OB is a sector of the circle with centre O.

Segment of a circle

Let PQ be a chord of a circle with centre O and radius r, then PQ divides the region enclosed by the circle into two parts. Each part is called a segment of the circle. The part containing the minor arc is called the minor segment and the part containing the major arc is called the major segment.

INTERSECTION OF A CIRCLE AND A LINE

Consider a circle with centre O and radius r and a line PQ in a plane. We find that there are three different positions a line can take with respect to the circle as given below in fig.

(a) The line PQ does not intersect the circle.
In fig. (a) the line PQ and the circle have no common point. In this case PQ is called a non-intersecting line with respect to the circle.
(b) The line PQ intersect the circle in more than one point. In fig. (b), there are two common points A and B between the line PQ and the circle and we call the line PQ as a secant of the circle.
(c) The line intersect the circle in a single point i.e. the line intersect the circle in only one point. In fig. (c) you can verify that there is only one point 'A' which is common to the line PQ in the given circle. In this case the line is called a tangent to the circle.

Secant

A secant is a straight line that cuts the circumference of the circle at two distinct (different) points i.e., if a circle and a line have two common points then the line is said to be secant to the circle.

Tangent

A tangent is a straight line that meets the circle at one and only one point. This point 'A' is called point of contact or point of tangency in fig. (c).

Tangent as a limiting case of a secant
In the fig. the secant cuts the circle at A and B. If this secant is turned around the point A, keeping A fixed then B moves on the circumference closer to A. In the limiting position, B coincides with A. The secant becomes the tangent at A. Tangent to a circle is a secant when the two end points of its corresponding chord coincide.

In the fig. is a secant which cuts the'circle at A and B. If the secant is moved parallel to itself away from the centre, then the points A and B come closer and closer to each other. In the limiting position, they coincide into a single point at A, the secant becomes the tangent at A. Thus a tangent line is the limiting case of a secant when the two points of intersection of the secant and a circle coincide with the point A. The point A is called the point of contact of the tangent. The line touches the circle at the point A. i.e., the common point of the tangent and the circle is called the point of contact and the tangent is said to touch the circle at the common point.

Note : The line containing the radius through the point of contact is called normal to the circle at the point.

NUMBER OF TANGENTS TO A CIRCLE FROM A POINT

1. If a point A lies inside a circle, no line passing through 'A' can be a tangent to the circle. i.e., No tangent can be drawn from the point A.
 

2. If A lies on the circle, then one and only one tangent can be drawn to pass through 'A'.

i.e. Exactly one tangent can be drawn through A.

3. If A lies outside the circle then exactly two tangents can be drawn through 'A'.

In the fig., a secant ABC is drawn from a point 'A' outside the circle, if the secant is turned around A in the clockwise direction, in the limiting position, it becomes a tangent at T. Similarly if the secant is turned tangent at S. Thus from a point A outside a circle only two tangents can be drawn. The points S and T where the lines touch the circle are called the points of contact.

PROPERTIES OF TANGENT TO A CIRCLE

Theorem-1 : The tangent at any point of a circle and the radius through the point are perpendicular to each other.

Given : A circle with centre O. AB is a tangent to the circle at a point P and OP is the radius through P.

To prove : OP AB.

Construct : Take a point Q, other than P, on tangent AB. Join OQ.

Proof :

Hence, proved.

Remark 1: A pair of tangents drawn at two points of a circle are either parallel or they intersect each other at a point outside the circle.
Remark 2: If two tangents drawn to a circle are parallel to each other, then the line-segment joining their points of contact is a diameter of the circle.
Remark 3: The distance between two parallel tangents to a circle is equal to the diameter of the circle, i.e.,
twice the radius.
Remark 4: A pair of tangents drawn to a circle at the end points of a diameter of a circle are parallel to each other.
Remark 5: A pair of tangents drawn to a circle at the end points of a chord of the circle, other than a diameter,
intersect each other at a point outside the circle.
Corollary 1: A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle
Given : O is the centre and r be the radius of the circle. OP is a radius of the circle. Line  is drawn through
P so that OP ⊥ ℓ

To prove : Line ℓ is tangent to the circle at P.

Construction : Suppose that the line is not the tangent to the circle at P. Let us draw another straight line
m which is tangent to the circle at P. Take two points A and B (other than P) on the line and two points
C and D on m.
Proof: 

But a part cannot, be equal to whole. This gives contradiction. Hence, our supposition is wrong.

Therefore, the line ℓ is tangent to the circle at P
Corollary 2 : If O be the centre of a circle and tangents drawn to the circle at the points A and B of the circle intersect each other at P, then ∠AOB + ∠APB = 180°.

Proof:

Hence, Proved

Theorem-2 : If two tangents are drawn to a circle from an exterior point, then
(i) the tangents are equal in length
(ii) the tangents subtend equal angles at the centre
(iii) the tangents are equally inclined to the line joining the point and the centre of the circle.

Given : PA and PB are two tangents drawn to a circle with centre O, from an exterior point P.

To prove : (i) PA = PB (ii) ∠AOP = ∠BOP, (iii) ∠APO = ∠BPO.

Proof :

Corollary 3 : If PA and PB are two tangents from a point to a circle with centre 0 touching it at A and B Prove that OP is perpendicular bisector of AB.

Proof:

Therefore, OP is perpendicular bisector of AB

Hence Proved.

COMMON TANGENTS OF TWO CIRCLES

Two circles in a plane, either intersect each other in two points or touch each other at a point or they neither intersect nor touch each other.

Common Tangent of two intersecting circles : Two circles intersect each other in two points A and B.

Here, PP' and QQ' are the only two common tangents. The case where the two circles are of unequal radii, we find the common tangents PP' and QQ' are not parallel.

Common tangents of two circles which touch each other externally at a point :

Two circles touch other externally at C.

Here, PP', QQ' and AB are the three common tangents drawn to the circles.

Common tangents of two circles which touch each other internally at a point :

Two circles touch other internally at C. Here, we have only one common tangent of the two circles.

Common tangents of two non-intersecting and non-touching circles :

Here, we observe that in figure (a), there is no common tangent but in figure (b) there are four common tangents PP', QQ', AA and BB'.

Ex.1 A point A is 26 cm away from the centre of a circle and the length of tangent drawn from A to the circle is 24 cm. Find the radius of the circle.

Sol. Let O be the centre of the circle and let A be a point outside the circle such that OA = 26 cm.

Let AT be the tangent to the circle.

Then, AT = 24 cm. Join OT.

Since the radius through the point of contact is perpendicular to the tangent, we have ∠OTA = 90°. In right ∠OTA, we have

OT² = OA² - AT²

= [(26)² - (24)²] = (26 + 24) (26 - 24) = 100.

Hence, the radius of the circle is 10 cm.

Ex.2 In the given figure, ΔABC is right-angled at B, in which AB = 15 cm and BC = 8 cm. A circle with centre O has been inscribed in ΔABC. Calculate the value of x, the radius of the inscribed circle.

Sol. Let the inscribed circle touch the sides AB, BC and CA at P, Q and R respectively. Applying Pythagoras theorem on right ΔABC, we have

AC² = AB² + BC² = (15)² + (8)² = (225 + 64) = 289

[∵ ∠OPB = 90°, ∠PBQ = 90°, ∠OQB = 90° and OP = OQ = x cm]

∵ BP = BQ = x cm.

Since the tangents to a circle from an exterior point are equal in length, we have AR = AP and CR = CQ.

Now, AR = AP = (AB – BP) = (15 – x) cm

CR = CQ = (BC – BQ) = (8 – x) cm.

∴ AC = AR + CR ⇒ 17 = (15 – x) + (8 – x) ⇒ 2x = 6 ⇒ x = 3.

Hence, the radius of the inscribed circle is 3 cm.

Ex.3 If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.

Sol. Let ABCD be a parallelogram whose sides AB, BC, CD and DA touch a circle at the points P, Q, R and S respectively.

Since the lengths of tangents drawn from an external point to a circle are equal, we have

AP = AS, BP = BQ, CR = CQ and DR = DS.

∴ AB + CD = AP + BP + CR + DR

= AS + BQ + CQ + DS
= (AS + DS) + (BQ + CQ)
= AD + BC

Now, AB + CD = AD + BC

⇒ 2AB = 2BC [ Opposite sides of a  gm are equal]
⇒ AB = BC
∴ AB = BC = CD = AD.

Hence, ABCD is a rhombus.

Ex.4 In the given figure, the incircle of ΔABC touches the sides AB, BC and CA at the points P, Q, R respectively.

Show that AP + BQ + CR = BP + CQ + AR = 1/2 (Perimeter of ΔABC)

Sol. Since the lengths of two tangents drawn from an external point to a circle are equal, we have

AP = AR, BQ = BP and CR = CQ
∴ AP + BQ + CR = AR + BP + CQ ...(i)

Perimeter of ΔABC = AB + BC + CA = AP + BP + BQ + CQ + AR + CR
= (AP + BQ + CR) + (BP + CQ + AR)
= 2(AP + BQ + CR) [Using (i)]

∴ AP + BQ + CR = BP + CQ + AR = 1/2 (Perimeter of ΔABC)

Ex.5 In two concentric circles, prove that a chord of larger circle which is tangent to smaller circle is bisected at the point of contact.

Sol. Let there be two concentric circles, each with centre O.

Since OP is a radius of smaller circle and APB is a tangent to it at the point P, so OP  AB.

But the perpendicular from the centre to a chord, bisects the chord.
∴ AP = PB
Hence, AB is bisected at the point P.

Ex.6 Two concentric circles are of radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.

Sol. Let O be the centre of the concentric circles and let AB be a chord of the outer circle, touching the inner circle at P. Join OA and OP.

Now, the radius through the point of contact is perpendicular to the tangent.
∴ OP ⊥ AB.

Since, the perpendicular from the centre to a chord, bisects the chord, AP = PB. Now, in right OPA, we have

OA = 13 cm and OP = 5 cm.
∴ OP² + AP2 = OA² ⇒ AP2 = OA² – OP² = (132 – 52) = (169 – 25) = 144.

∴ AB = 2AP = (2 x 12) cm = 24 cm.
Hence, the lenath of chord. AB = 24 cm.

Ex.7 In the given figure, PT is a common tangent to the circles touching externally at P and AB is another common tangent touching the circles at A and B. Prove that:
(i) T is the mid-point of AB
(ii) ∠APB = 90°
(iii) If X and Y are centres of the two circles, show that the circle on AB as diameter touches the line XY.

Sol. (i) Since the two tangents to a circle from an external point are equal, we have

TA = TP and TB = TP.
∴ TA = TB [Each equal to TP]

(i) 

Hence, T bisects AB, i.e., T is the mid-point of AB.
TA = TP ⇒ ∠TAP = ∠TPA
TB = TP ⇒ ∠TBP = ∠TPB
∴ ∠TAP + ∠TBP = ∠TPA + ∠TPB = ∠APB
⇒ ∠TAP + ∠TBP + ∠APB = 2∠APB
⇒ 2∠APB = 180°    [∴The sum of the ∠s of a Δ is 180°]
⇒ ∠APB = 90°

(ii) Thus, P lies on the semi-circle with AB as diameter.

Hence, the circle on AD as diameter touches the line XY.

Ex.8 Two circles of radii 25cm and 9cm touch each other externally. Find the length of the direct common tangent.

Sol. Let the two circles with centres A and B and radii 25 cm and 9 cm respectively touch each other externally
at a point C.
Then, AB = AC + CB = (25 + 9) cm = 34 cm.
Let PQ be a direct common tangent to the two circles.
Join AP and BQ.
Then, AP ⊥ PQ and BQ ⊥ PQ.
[∴ Radius through point of contact is perpendicular to the tangent]

Draw, BL ⊥ AP.

Then, PLBQ is a rectangle.
Now, LP = BQ = 9 cm and PQ = BL.
∴ AL = (AP – LP) = (25 – 9) cm = 16 cm.
From right ΔALB, we have

AB² = AL² + BL² ⇒ BL² = AB² – AL² = (34)² – (16)² = (34 + 16) (34 – 16) = 900

∴ PQ = BL = 30 cm.
Hence, the length of direct common tangent is 30 cm.

Ex.9 In the given figure, PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of : (i) ∠QOP (ii) ∠QCP

Sol. (i) In ΔPQR,
PQ = QR ⇒ ∠PRQ = ∠QPR                 [∠s opp. to equal sides of a Δ are equal]
Also, ∠QPR + ∠RQP + ∠PRQ = 180°        [Sum of the ∠s of a Δ is 180°]
⇒ 68° + 2∠PRQ = 180°
⇒ 2∠PRQ = (180° – 68°) = 112°
⇒ ∠PRQ = 56°.
∴ ∠QOP = 2∠PRQ = (2 × 56°) = 112°.        [Angle at the centre is double the angle on the circle]
(ii) Since the radius through the point of contact is perpendicular to the tangent, we have
∠OQC = 90° and ∠OPC = 90°.

Now, ∠OQC + ∠QOP + ∠OPC + ∠QCP = 360° [Sum of the ∠s of a quad. is 360°)
⇒ 90° + 112° + 90° + ∠QCP = 360°.
⇒ ∠QCP = (360° – 292°) = 68°.

Ex.10 With the vertices of ΔABC as centres, three circles are described, each touching the other two externally. If the sides of the triangle are 9 cm, 7 cm and 6 cm, find the radii of the circles.

Sol. Let AB = 9 cm, BC = 7 cm and CA = 6 cm.
Let x, y, z be the radii of circles with centres A, B, C respectively.

Then, x + y = 9, y + z = 7 and z + x = 6.

Adding, we get 2(x + y+ z) = 22 ⇒ x + y + z = 11.

∴ x = [(x + y + z) - (y + z)] = (11 - 7) cm = 4 cm.

Similarly, y = (11-6) cm = 5 cm and z = (11-9) cm = 2 cm.

Hence, the radii of circles with centres A, B, C are 4 cm, 5cm and 2 cm respectively.