Introduction:
We know 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 circle like chord, segment, sector, etc. Let us now see, how many different positions a line can take with respect to a given circle C (O, r). Let the line be 'l'.
In figure (i) the line 'l' does not touch or intersect the circle.
In figure (ii) the line 'l' touch the circle in only one point. In this case the line is said to be a tangent to the circle and point is called the point of contact to the tangent.
Tangent: A tangent to a circle is a line that intersects the circle in exactly one point.
* The word 'Tangent' is originated from the Latin word 'TANGERE' which means 'to touch'.
In figure (iii) the line 'l', intersects the circle in two distinct points. In such a case the line 'l' is called a 'secant' of the circle.
Secant: A line which intersects a circle in two distinct points is called a secant of the circle.
Some properties of tangent to a circle:
We shall now prove some properties of tangents to a circle as theorems.
Theorem 1: A tangent to a circle is perpendicular to the radius through the point of contact.
Proof:
Given: A circle C (O, r) and a tangent AB at point 'P'.
Required to prove (RTP): OP ⊥ AB
Construction: Take any point Q, other than 'P', on the tangent AB. Join OQ, suppose OQ meets the circle at R.
Proof: We know that among all line segments joining the point 'O' to a point on AB, the shortest one is perpendicular to AB. So, to prove that OP ⊥ AB, it is sufficient to prove that OP is shorten than any other segment joining 'O' to any point of AB.
It is obvious that OP = OR (Radii of same circle)
Now OQ = OR + RQ
⇒ OQ > OR
⇒ OQ > OP
⇒ OP < OQ
Thus, OP is shorter than any other segment joining to any point of AB.
Hence OP ⊥ AB.
Theorem 2: (Converse of the Theorem 1)
A line drawn through the end point of radius and perpendicular to it is a tangent to the circle.
Proof:
Given: A radius OP of a circle C (O, r) and a line APB, perpendicular to OP.
RTP: AB is a tangent to the circle at the point 'P'.
Proof: Take a point 'Q' different from 'P' on the line AB
Now OP ⊥ AB
⇒ Among all the line segments joining 'O' to a point on AB, OP is the shortest.
⇒ OP < OQ
⇒ OQ > OP
⇒ Q lies out side the circle
Thus, every point on AB, other than P, lies outside the circle. This shows that AB meets the circle only at the point P.
Hence, AB is a tangent to the circle at 'P'.
Construction of tangent to a circle
I. Here we construct a tangent to a circle at a given point when the centre of the circle is known.
Construct a tangent of a circle having radius 3 cm with centre 'O'.
Steps of construction:
1. Draw a circle of radius 3 cm with centre 'O' and make a point 'P' anywhere on it. Joint OP.
2. Extend OP up to Q.
3. Identify two points A and B on OQ such that PA = PB.
(with centre 'P' draw two equal length arcs on OQ)
4. Draw perpendicular bisector XY to AB.
5. XY is the required tangent to the given circle passing through 'P'.
Note: We can draw infinite number of tangents to a circle.
II. Construction of tangent to a circle at a given point when its centre is not known.
Steps of construction:
1. Draw a circle of given radius.
2. Draw any chord PQ through the given point 'P' on the circle.
3. Join P and Q to a point R either in the major arc or in the minor arc.
4. Construct ∠QPY equal to ∠PRQ and on the opposite side of the chord PQ.
5. Produce YP to X to get YPX as the required tangent.
Finding the length of the tangent:
Examples
1. Find the length of the tangent to a circle with radius 6 cm from a point 'P' such that OP = 10 cm.
Sol: We know that tangent is perpendicular to the radius at the point of contact.
Here PA is tangent segment and OA is radius of circle.
∴ OA ⊥ PA ⇒ ∠OAP = 90°
Now in Δ OAP, OP2 = OA2 + PA2 (Pythagoras theorem)
⇒ 102 = 62 + PA2
⇒ 100 = 36 + PA2
⇒ PA2 = 100 − 36
⇒ PA2 = 64
∴ PA = 
= 8 cm
2. 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 = 13 cm. Find the length of PQ.
Sol: We know that tangent is perpendicular to the radius at the point of contact. Here PQ is tangent segment and OP is radius of circle.
∴ OP ⊥ PQ ⇒ ∠OPQ = 90°
Now in Δ OPQ, OQ2 = OP2 + PQ2
132 = 52 + PQ2
⇒ PQ2 = 169 − 25
⇒ PQ = 
= 12 cm
3. Prove that the tangents to a circle at the end points of a diameter are parallel.
Sol: Let 'O' is the centre of the circle and AB is diameter. OA is radius, PAQ is tangent.
∠OAQ = 90°
(... Angle between radius & tangent)
Similarly OB is radius, RBS is tangent ∠OBR = 90°
(∵ Angle between radius & tangent)
But ∠OAQ, ∠OBR are alternate angles
∴ PQ // RS
Number of tangents to a circle from any point
i) There is no tangent to a circle passing through a point lying inside the circle.
ii) There is one and only one tangent to a circle passing through a point lying on the circle.
iii) There are exactly two tangents to a circle through a point lying outside the circle in this case, A and B are the points of contacts of the tangents PA and PB respectively.
Theorem 3: The lengths of tangents drawn from an external point to a circle are equal.
Given: A circle with centre 'O', 'P' is a point lying outside the circle and PA and PB are two tangents to the circle from 'P'.
R.T.P: PA = PB
Proof: Join OA, OB and OP.
∠OAP = ∠OBP = 90°
(∵ Angle between radii and tangents)
Now in the two right triangles
ΔOAP and ΔOBP
OA = OB (radii of same circle)
OP = OP (Common)
∴ By RHS congruency axiom,
ΔOAP ≅ ΔOBP
⇒ PA = PB
Hence Proved.
Construction of Tangents to a circle from an external point
Given: 'O' is the centre of the circle 'P' is a point outside it.
Steps of construction:
1. Joint PO and draw perpendicular bisector to it. Let M be the mid point of PO.
2. Taking M as centre and PM or MO as radius, draw a circle. Let it intersect the given circle at the points A and B.
3. Join PA and PB. Then PA and PB are the required two tangents.
Proof: Join OA, then ∠PAO is an angle in the semi circle and
∠PAO = 90°
∴ PA ⊥ OA
Since, OA is radius of the given circle, PA has to be a tangent to the circle similarly, PB is also tangent to the circle.
Hence Proved.
Some interesting statements about tangents and secants and their Proof.
Statement − 1: The centre of a circle lies on the bisector of the angle between two tangents drawn from a point outside it.
Proof: Let PQ and PR be two tangents drawn from a point 'P' outside of the circle with centre 'O'.
Join OQ and OR, triangles OQP and ORP are congruent because we know that,
∠OQP = ∠ORP = 90°
OQ = OR (radii)
OP = OP (common)
⇒∠OPQ = ∠OPR
∴ OP is the angle bisector of ∠QPR
Hence, the centre lies on the bisector of the angle between the two tangents.
Statement − 2: In two concentric circles, such that a chord of the bigger circle, that touches the smaller circle is bisected at the point of contact with the smaller circle.
Proof: We have two concentric circles C1 and C2 with centre O and a chord AB of larger circle C1, touching the smaller circle C2 at the point P.
We have to prove that AP = PB
Join OP
Then AB is a tangent to the circle C2 at P and OP is its radius.
∴ OP ⊥ AB
Now Δ OAP ≅ Δ OBP (RHS congruency) ⇒ AP = PB.
∴ OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord.
Statement 3: If two tangents AP and AQ are drawn to a circle with centre 'O' from an external point A then ∠PAQ = 2 ∠OPQ = 2 ∠OQP.
Proof: We are given a circle with centre 'O' an external point A and two tangents AP and AQ to the circle, where P, Q are the points of contact.
∠PAQ = 2 ∠OPQ
Let ∠PAQ = θ
AP = AQ, So ΔAPQ is an isosceles triangle
∴ ∠APQ + ∠AQP + ∠PAQ = 180°
∠APQ = ∠AQP = 
(180° − θ)
= 90° − θ
⇒∠OPA = 90°
So, ∠OPQ = ∠OPA − ∠APQ
= 90° − [90° − θ]
= θ
= ∠PAQ
∴ ∠PAQ = 2 ∠OPQ
Similarly ∠PAQ = 2 ∠OQP.
Statement 4: If a circle touches all the four sides of a quadrilateral ABCD at points PQRS. Then AB + CD = BC + DA.
Proof: The circle touched the sides
AB, BC, CD and DA of quadrilateral ABCD at the points P, Q, R and S respectively as shown.
We know that the two tangents to a circle drawn from a point outside it, are equal
AP = AS
BP = BQ
DR = DS
CR = CQ
on adding, we get
AP + BP + DR + CR = AS + BQ + DS + CQ
⇒ (AP + PB) + (CR + DR) = (BQ + QC) + (DS + SA)
⇒ AB + CD = BC + DA
Segment of a circle formed by a secant
In the figure, we have seen a line and a circle. When a line meets a circle in only one point, it is tangent. A secant is a line which intersects the circle at two distinct points represented in the chord. Here 'l' is the secant and AB is the chord.
Finding the area of segment of a circle
As we know a segment is a region, bounded by the arc and a chord, we can see the area that is shaded part is a minor segment (APB), semicircle in unshaded part is major segment (AQB).
How do we find the area of the segment?
The portion of some unshaded part (major segment) and shaded part (minor segment) is a sector which is the combination of a triangle and a segment.
Let OAPB be a sector of a circle with centre 'O' and radius 'r' as shown in the figure.
4. Find the area of the segment AYB showing in the adjacent figure. If radius of the circle is 21 cm and ∠AOB = 120°.
Sol: Area of the segment AYB
= Area of sector OAYB − Area of ΔOAB
= 462 cm2 ............... (1)
finding the area of ΔOAB, draw OM ⊥ AB
we know that OA = OB
by RHS congruency ΔAMO ≅ ΔBMO
So, M is the mid point of AB and ∠AOM = ∠BOM = × 120° = 60°
Let OM = x cm
5. Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm each (use Π = 3.14).
Sol: Let us assume that the two unshaded regions as I and II.
radius = 10 cm
= 157 cm2 ............... (1)
But area of ABCD square = 10 cm × 10 cm
= 100 cm2
∴ I + II + shaded region = 100 cm2
⇒ I + II + shaded region + shaded region = 157 cm2
⇒ 100 cm2 + shaded region = 157 cm2
⇒ shaded region = 157 - 100 = 57 cm2
Writer: T.S.V.S. Surya Narayanamurthy
