**Tangents**** of a Circle**

A line which is drawn from an external
point of a circle and meets the circle at one and only point on the
circumference is called a **tangent of a circle**.

The point A where tangent touches the
circle is called the **point of contact**.

In the figure, TAN is a line which meets
the circle at a point A. So TAN is a **tangent** and the point A is **point of contac****t**.

In the same figure, a line TB cuts the
circle at two points D and E. So the line TB is called a **secant**.

**Properties of tangent:**

1.
A tangent to a circle is
perpendicular to the radius of the circle drawn at point of contact.

2. The length of two tangents drawn from an external point to a circle are always equal.

**Angles in alternate segments**

In the figure given below, PQ is a tangent to a
circle at the point of contact T. A chord TA is drawn from the point of contact
T. Here the chord TA divides the circle into two segments, TBA and TCA. These
two segments are called alternate segments.

Here, for ∠ATP, the alternate segment is TBA.

And, for ∠ATQ, the alternate segment is TCA.

So, ∠ATP and ∠TBA are said to be the alternate segment angles and, ∠ATQ and ∠TCA are also said to be the
alternate segment angles.

The alternate segment angles are always
equal.

i.e ∠ATP = ∠TBA and ∠ATQ = ∠TCA

**Circle theorems on tangents:**

1.
A tangent to a circle is
perpendicular to the radius of the circle drawn at point of contact.

2.
The length of two tangents
drawn from an external point to a circle are always equal.

3.
The alternate segment angles
are always equal.

**Proofs:**

**1. **__A tangent to
a circle is perpendicular to the radius of the circle drawn at point of contact__**.**

__Given__**:** O is centre of the circle.
PQ is a tangent to circle at point of contact T.

__To Prove__**:** OT ⊥ PQ

__Construction__**:** Any point R is taken on PQ
and OR is joined.

__Proof__**:**

__Statements____Reasons__

1.
OT = OS -----------------> Radii of
the same circle.

2.
OS < OR -----------------> OS is
a part of OR.

3.
OT < OR -----------------> From
statements 1 and 2.

4.
OT is the shortest length of
all the lines drawn from O to PQ ------> It is true for every line joining O
to PQ.

5.
OT ⊥ PQ ------> Perpendicular
is the shortest line segment drawn from a point to a line.

Proved.

**2. **__The length
of two tangents drawn from an external point to a circle are always equal__**.**

__Given__**:** O is the centre of a circle. PQ and PR are two tangents from the same point P to the circle at T1 and T2.

__To Prove__**:** PT_{1} = PT_{2}

__Construction__**: **OP, OT_{1} and OT_{2}
are joined.

__Proof__**:**

__Statements____Reasons__

1.
In DOPT_{1} and DOPT_{2}

i. ∠OT_{1}P = ∠OT_{2}P (R) -------> Both of them are right angles.

ii. OP = OP (H)
---------------> Common side.

iii. OT_{1} = OT_{2} (S) ------------> Radii of the same circle.

2.
DOPT_{1} ≅ DOPT_{2} ----------------> By RHS axiom.

3.
PT_{1} = PT_{2}
---------> Corresponding sides of congruent triangles.

Proved.

**3. **__The alternate
segment angles are always equal__**.**

__Given__**:** O is the centre of a circle.
PQ is a tangent to the circle at point of contact T. TA is a chord. ∠ATP and ∠TBA; ∠ATQ and ∠TCA are two pair of alternate
segment angles.

__To Prove__**:** ∠ATP = ∠TBA and ∠ATQ = ∠TCA

__Construction__**: **OT joined and TO is produced
to R, a point on the circumference. Hence, TR is the diameter of the circle. RA
is joined.

__Proof__**:**

__Statements____Reasons__

1.
∠RAT = 90° --------------> Angle in the semi-circle.

2.
∠RTA + ∠TRA = 90° -----> Sum of two acute angles of a rt. angled DRAT.

3.
∠RTA + ∠ATQ = 90° -----> RT ⊥ PQ

4.
∠RTA + ∠TRA = ∠RTA + ∠ATQ ------> From statements 2 and 3.

i.e. ∠TRA = ∠ATQ

5.
∠TRA = ∠TCA ----------> Inscribed angles on same arc ABT.

6.
∠TCA = ∠ATQ ------------> From statements 4 and 5.

7.
Similarly, ∠ATP = ∠TBA ---------> Same facts and
reasons as above.

Proved.

Look at the following worked out examples
on tangents of a circle:

*Workout Examples*

*Workout Examples*

*Example 1: In the given figure, O is the centre of circle**. TAN is a tangent to the circle where A
is the point of contact. If **∠**BAN = 40° and **∠**ABO = x°, find the value of x.*

*Solution:** Here,*

* **∠**BAN = 40°*

* **∠**OAN = 90° [**∵** OA **⊥** TN]*

* **∠**OAN
= **∠**OAB + **∠**BAN [**∵** Whole part axiom]*

* i.e. 90° = **∠**OAB
+ 40°*

* or, **∠**OAB
= 90° - 40°*

* or, **∠**OAB
= 50°*

* **∠**ABO
= **∠**OAB = 50° [**∵** OA = OB]*

* ∴** x
= 50°*

*Example 2: In the given fugure, O is the centre of circle**. TAN is a tangent to the circle at A
where A is the point of contact. If OA = 8cm and AT = 15cm, find the length of
BT.*

*Solution:** Here,*

* AT
= 15cm, OA = 8cm, BT = ?*

* **∠**OAT = 90° [**∵** OA **⊥** TN]*

* **∴** **D**OAT
is a right angled triangle*

* OT ^{2}
= OA^{2} + AT^{2} [*

*∴*

**Pythagoras****relation]*** i.e. OT ^{2} = 8^{2} + 15^{2}*

* or, OT ^{2} = 64 + 225*

* or, OT ^{2} = 289*

* or, OT = √289 = 17cm*

* Now,*

* BT = OT – OB*

*
= OT – OA [**∵** OA = OB, radii of same circle]*

*
= 17cm – 8cm*

*
= 9cm*

*Example 3: In the given figure, O is the centre of a circle**. TAN is a tangent to the circle at A. If
**∠**ACB = 65° and **∠**CAT = x°, find the value of x.*

*Solution:** From the figure,*

* **∠**CAB
= 90° [**∵** Inscribed angle on semi-circle]*

* **∠**CAB
+ **∠**ACB + **∠**ABC = 180° [Sum of angles of **D**ABC]*

* i.e. 90° + 65° + **∠**ABC = 180°*

* or, 155° + **∠**ABC
= 180°*

* or, **∠**ABC
= 180° - 155°*

* or, **∠**ABC
= 25°*

* *

* **∠**CAT
= **∠**ABC [**∵** Angles in the alternate segments]*

* **∴**
x = 25°*

*Example 4: In the adjoining figure, two circles intersect each other
at A and D. Their common tangent TP meets the two circles at B and C. Prove
that, **∠**BAC + **∠**BDC = two right angles.*

*Solution: *

__Given__*:
Two circles intersect each other at A and D. TP is a common tangent at B and C.*

__To Prove__*:
**∠**BAC + **∠**BDC = 180°*

__Construction__*: Points A and D are joined.*

__Proof__*:*

* Statements Reasons*

*1.
**∠**DAB = **∠**DBC -----> Being angles in the alternate segment of the circle.*

*2.
**∠**DAC = **∠**DCB -----> By reason similar to 1.*

*3.
**∠**DAB + **∠**DAC = **∠**DBC + **∠**DCB -----> By adding statements 1 and 2.*

*i.e. **∠**BAC = **∠**DBC + **∠**DCB*

*4.
**∠**DBC + **∠**DCB + **∠**BDC = 180° ------> Sum of angles of **D**DBC.*

*5.
**∠**BAC + **∠**BDC = 180° ------> From statements 3 and 4.*

* Proved.*

*Example 5: In the given figure, PQ is a tangent to the circle at point of contact A.
C is the middle point of arc AB. If CM **⊥** PQ and CN **⊥** AB, prove that CM = CN.*

*Solution: *

__Given__*:
O is the centre of circle. PQ is a tangent to the circle at point of contact A. C is the
mid-point of arc ACB i.e. arc AC = arc BC, CM **⊥** PQ and CN **⊥** AB.*

__To Prove__*:
CM = CN*

__Construction__*: AC and BC joined.*

__Proof__*:*

* Statements Reasons*

*1.
**Chord
AC = chord BC ---------> being arc AC = arc BC.*

*2.
**∠**BAC = **∠**ABC --------------> Being AC = BC in **D**ABC.*

*3.
**∠**CAM = **∠**ABC ---------------> Alternate segment angles of a circle.*

*4.
**∠**BAC = **∠**CAM ----------------> From statements 2 and 3.*

*5.
**In
**D**ACM and **D**ACN*

*i. **∠**AMC = **∠**ANC (A) --------> Both being 90°.*

*ii. **∠**CAM = **∠**CAN (A) -------> From statement 4.*

*iii. **AC
= AC (S) -------> Common side.*

*6.
**D**ACM **≅** **D**ACN -----------> By AAS axiom.*

*7.
**CM = CN** -------------> Corresponding
sides of **≅** triangles.*

* Proved.*

*You can comment your
questions or problems regarding circle and tangents here.*

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