Table of contents  
Introduction to Circles  
Tangent to a Circle  
Number of Tangents from a Point on a Circle  
Summary 
Circle: It is a closed twodimensional geometrical figure, such that all points on the surface of a circle are equidistant from the point called the “centre”.
A circle is a locus of a point that moves in such a way that the distance from that point is always fixed.
Parts of a Circle
Nonintersecting lines: These are made up of two or more lines that do not intersect. The circle and the line AB have no common point.
 Lines that do not intersect can never meet.
 The parallel lines are another name for them.
 They stay at the same distance from one another at all times.
A tangent to a circle is a line that intersects the circle at only one point.
(i) There is only one tangent at a point of the circle.
(ii) The tangent to a circle is a special case of the secant when the two endpoints of its corresponding chord coincide.
(iii) 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 that point.
The theorem states that “the tangent to the circle at any point is perpendicular to the radius of the circle that passes through the point of contact”.
According to the theorem, O is the centre and OP⊥XY.
Note:
1. By the theorem above, we can also conclude that at any point on a circle, there can be one and only one tangent.
2. The line containing the radius through the point of contact is also sometimes called the ‘normal’ to the circle at the point.
Example 1: In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100°, then calculate ∠BAT.
Solution:
∠1 = ∠2
∠1 + ∠2 + 100° = 180°
∠1 + ∠1 = 80°
⇒ 2∠1 = 80°
⇒ ∠1 = 40°
∠1 + ∠BAT = 90°
∠BAT = 90° – 40° = 50°
1. There is no tangent to a circle passing through a point lying inside the circle.
If a point lies inside a circle, any line passing through that point will intersect the circle at two points and is called a secant. Therefore, it is not possible to draw a tangent to a circle that passes through a point inside the circle.
2. There is one and only one tangent to a circle passing through a point lying on the circle.
When a point lies on the circle, there is exactly one tangent to a circle that passes through it.
3. There are exactly two tangents to a circle through a point lying outside the circle
If a point is located outside of a circle, then there exist exactly two tangents that can be drawn to the circle passing through the point.
To define the length of a tangent from a point (P) to a circle, we measure the distance from the external point P to the point of tangency "I" on the circle. This distance is known as the tangent length (PI).
The lengths of tangents drawn from an external point to a circle are equal.
Note:
1. The theorem can also be proved by using the Pythagoras Theorem as follows:
PA^{2} = OP^{2} – OA^{2} = OP^{2} – OB^{2} = PB^{2} (As OA = OB) which gives PA = PB.
2. Note also that ∠ OPA = ∠ OPB. Therefore, OP is the angle bisector of ∠ APB,i.e., the centre lies on the bisector of the angle between the two tangents.
Example 2: In the given figure, AB and AC are tangents to the circle with centre o such that ∠BAC = 40°. Then calculate ∠BOC.
Solution:
AB and AC are tangents
∴ ∠ABO = ∠ACO = 90°
In ABOC,
∠ABO + ∠ACO + ∠BAC + ∠BOC = 360°
90° + 90° + 40° + ∠BOC = 360°
∠BOC = 360 – 220° = 140°
Example 3: In the given figure, PA and PB are tangents to the circle with centre O. If ∠APB = 60°, then calculate ∠OAB.
Solution:
∠1 = ∠2
∠1 + ∠2 + ∠APB = 180°
∠1 + ∠1 + 60° = 180°
2∠1 = 180° – 60° = 120°
1
∠1 = 60°
∠1 + ∠OAB = 90°
60° +∠OAB = 90°
∠OAB = 90° – 60° = 30°
Example 4: In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70°, then calculate ∠TP.
Solution:
∠1 = ∠2
∠1 + ∠2 + 70° = 180°
∠1 + ∠1 = 180° – 70°
2∠1 = 110° ⇒ ∠1 = 55°
∠1 + ∠TPQ = 90°
55° + ∠TPQ = 90°
⇒ ∠TPQ = 90° – 55° = 35°
Example 5: In the given figure, PQ R is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA.
Solution:
∠ACB = 90° …[Angle in the semicircle
In ∆ABC,
∠CAB + ∠ACB + ∠CBA = 180°
30 + 90° + ∠CBA = 180°
∠CBA = 180° – 30° – 90° = 60°
∠PCA = ∠CBA …[Angle in the alternate segment
∴ ∠PCA = 60°
124 videos457 docs77 tests

1. How many tangents can be drawn from a point outside a circle? 
2. Can a circle have a tangent at more than one point? 
3. Is the tangent to a circle always perpendicular to the radius at the point of contact? 
4. How does the number of tangents from a point on a circle change based on the position of the point? 
5. Can a line passing through the center of a circle be considered as a tangent to the circle? 

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