In our day-to-day real life situations, we have seen two lines intersect at a point or do not intersect in a plane. For example, two parallel lines in a railway track, do not intersect. Whereas, grills in a window intersect.

**Circles and Tangents**

In our day-to-day real life situations, we have seen two lines
intersect at a point or do not intersect in a plane. For example, two parallel
lines in a railway track, do not intersect. Whereas, grills in a window
intersect.

Similarly what happens when a curve and a line is given in a
plane? The curve may be parabola, circle or any general curve.

Similarly, what happens when we consider intersection of a line
and a circle?

**We may get three situations as given in the following diagram**

**Note**

The line segment *AB* inscribed in the circle in
Fig.4.52(c) is called **chord of**** ****the circle**. Thus a chord is a
sub-section of a secant.

**Definition**

**If a line touches the
given circle at only one point, then it is called ****tangent to the circle.**

(i) When a cycle moves along a road, then the road becomes the
tangent at each point when the wheels rolls on it.

(ii) When a stone is tied at one end of a string and is rotated
from the other end, then the stone will describe a circle. If we suddenly stop
the motion, the stone will go in a direction tangential to the circular motion.

** **

1. A tangent at any point on a circle and the radius through the
point are perpendicular to each other.

2. (a) No tangent can be drawn from an interior point of the
circle.

(b) Only one tangent can be drawn at any point on a circle.

(c) Two tangents can be drawn from any exterior point of a circle.

3. The lengths of the two tangents drawn from an exterior point to
a circle are equal,

**Proof : **By 1.** ***OA*** **┴** ***PA*,*OB*** **┴**
***PB*** **. Also** ***OA*=*OB*=** **radius,** **

*OP *is common side.* *∠*AOP *=* *∠*BOP*

Therefore, Δ*OAP *≅ Δ*OBP *. Hence *PA*
= *PB*

4. If two circles touch externally the distance between their
centers is equal to the sum of their radii, that is *OP* = *r*_{1}
+ *r*_{2}

Let two circles with centers at** ***O*** **and** ***P***
**touch other at** ***Q*.

Let *OQ* = *r*_{1} and *PQ* = *r*_{2}
and let *r*_{1} > *r*_{2} .

The distance between their centers *OP* = *d* . It is
clear from the Fig. 4.57 that when the circles touch externally *OP* = *d*
= *OQ* + *PQ* = *r*_{1} + *r*_{2} .

5. If two circles touch internally, the distance between their
centers is equal to the difference of their radii, that is *OP* = *r*_{1}
−*r*_{2} .

**Proof : **Let two circles with centers at** ***O*** **and** ***P***
**touch each other at** ***Q*. Let *OQ* = *r*_{1}
and *PQ* = *r*_{2} and let *r*_{1} > *r*_{2}
.

The distance between their centers *OP* = *d* . It is
clear from the Fig. 4.58 that when the circles touch internally, *OP* = *d*
= *OQ* – *PQ *

*OP *=* r*_{1}* *−*r*_{2}* *.

6. The two direct common tangents drawn to the circles are equal
in length, that is *AB* = *CD.*

The lengths of tangents drawn from *P* to the two circles are
equal.^{}

Therefore, *PA *=* PC *and* PB *=* PD. ^{}*

gives *PA* − *PB* = *PC* – *PD*

*AB *=* CD*

In the Fig. 4.60, the chord PQ divides the circle into two
segments. The tangent AB is drawn such that it touches the circle at *P*.

The angle in the alternate segment for ∠*QPB* (∠1) is ∠*QSP* (∠1) and that for ∠*QPA* (∠2) is ∠*PTQ* (∠2) .

If a line touches a circle and from the point of contact a chord
is drawn, the angles between the tangent and the chord are respectively equal
to the angles in the corresponding alternate segments.

**Given :** A circle with centre at *O*, tangent *AB* touches the circle at *P*
and *PQ* is a chord. *S* and *T* are two points on the circle in
the opposite sides of chord *PQ*.

**To prove :** (i) ∠*QPB* =
∠*PSQ* and (ii) ∠*QPA* = ∠*PTQ*

**Construction:** Draw the diameter *POR*. Draw *QR*, *QS* and *PS*.

**Example 4.24**

Find the length of the tangent drawn from a point whose distance
from the**
**centre of a circle is 5
cm and radius of the circle is 3 cm.

*Solution*

Given OP = 5 cm, radius *r* = 3 cm

To find the length of tangent *PT*.

In right angled Δ*OTP,*

*OP*^{2}* *=* OT*^{2}* *+* PT*^{2}* *(by
Pythagoras theorem)

5^{2} = 3^{2} + *PT* ^{2}
gives *PT *^{2}* *=* *25* *−* *9*
*=* *16

Length of the tangent *PT *=* *4* *cm

PQ is a chord of length 8 cm to a circle of radius 5 cm. The
tangents at P** **and Q intersect at a point T. Find the length of the tangent TP.

Let *TR* = *y* . Since, OT is perpendicular bisector of
PQ.

*PR *=* QR *=* *4* *cm

In Δ*ORP*, *OP* ^{2} = *OR*^{2}
+ *PR*^{2}

*OR *^{2}* *=* OP*^{2}* *−* PR*^{2}

*OR*^{2}* *=* *5^{2}* *−* *4^{2}* *=* *25*
*−16* *=* *9* *⇒* OR *=* *3* *cm

*OT *=* OR*+*RT *= 3+*y *... (1)

In Δ*PRT* , *TP*^{2} = *TR*^{2}
+ *PR*^{2 }… (2)

and Δ*OPT* we have, *OT*^{2} = *TP*^{2
}+*OP*^{2}

*OT*^{2}* *=* *(*TR*^{2}* *+* PR*^{2}*
*)* *+*OP*^{2 }(substitute
for *TP*^{2} from (2))

(3 + *y* )^{2 }= *y*^{2} + 4 ^{2}
+ 5^{2 }(substitute for *OT*
from (1))

9 + 6*y* + *y*^{2 }= *y* ^{2} + 16
+ 25 Therefore *y* = *TR* = 16/3

6*y* = 41 − 9 we get *y* = 16/3

From (2), *TP*^{2} = *TR*^{2} + *PR*^{2}

**Example 4.26**

In Fig.4.64,** ***O*** **is the centre of a circle.** ***PQ*** **is a** **chord and the tangent *PR* at *P*
makes an angle of 50° with *PQ*. Find ∠*POQ* .

*Solution*

∠*OPQ*** **=

*OP *=* OQ *(Radii of a circle are equal)

∠*OPQ *=* *∠*OQP *=* *40° (Δ*OPQ* is isosceles)^{ }

∠*POQ *=* *180° −* *∠*OPQ *−* *∠*OQP*

∠*POQ* = 180° − 40° − 40° =
100°

**Example 4.27**

In Fig.4.65,** **Δ*ABC*** **is circumscribing a circle. Find the length of *BC.*

*Solution*

*AN *=* AM *=* *3 cm (Tangents drawn from same external point
are equal)

*BN *=* BL *= 4* *cm

*CL *=* CM *=* AC *–* AM *= 9* *–* *3 = 6* *cm

Gives *BC *=* BL *+* CL *=* *4* *+* *6* *=* *10* *cm

If radii of two concentric circles are** **4** **cm and** **5** **cm then find the length
of** **the chord of one circle
which is a tangent to the other circle.

*OA *= 4* *cm, OB* *= 5* *cm; also* OA *┴* BC *.

*OB *^{2}* *=* OA*^{2}* *+* AB*^{2}

5^{2} = 4^{2} + *AB*^{2 }gives *AB*^{2}
= 9

Therefore *AB* = 3 cm

*BC *=* *2AB hence* BC *=* *2* *×3* *= 6* *cm

Now let us discuss how to draw

(i) a tangent to a circle using its centre

(ii) a tangent to a circle using alternate segment theorem

(iii) pair of tangents from an external point

**Construction of a tangent to a circle (Using the centre)**

Draw a circle of radius** **3** **cm. Take a point** ***P*** **on this circle and draw
a tangent at *P*.

Given, radius *r* = 3 cm

Step 1: Draw a circle with centre at *O *of radius* *3*
*cm.

Step 2: Take a point *P* on the circle. Join *OP*.

Step 3: Draw perpendicular
line TT’ to *OP*
which passes through *P*.

Step 4: *TT* ^{‘} is the required tangent.

**Construct of a tangent to a circle (Using alternate segment
theorem)**

Draw a circle of radius** **4** **cm. At a point** ***L*** **on it draw a tangent to the circle** **using the alternate
segment.

Given, radius=4 cm

Step 1 : With *O* as the centre, draw a circle of radius 4
cm.

Step 2: Take a point *L* on the circle. Through *L* draw any chord *LM*.

Step 3: Take a point *M* distinct from *L *and *N*
on the circle, so that *L*, *M *and* N *are in
anti-clockwise direction. Join *LN*
and *NM.*

Step 4: Through *L* draw a tangent *TT *’ such that ∠*TLM* = ∠*MNL.*

Step 5:* TT* ‘ is the required tangent.

**Construction of pair of tangents to a circle from an external
point ***P***.**

Draw a circle of diameter 6 cm from a point** ***P*, which is** **8** **cm away** **from its centre. Draw
the two tangents *PA* and *PB* to the circle and measure their lengths.

Given, diameter (*d*) = 6 cm, we find radius (*r*) = 6/2
= 3 cm

Step 1: With centre at *O*, draw a circle of radius 3 cm.

Step 2: Draw a line *OP* of length 8 cm.

Step 3: Draw a perpendicular bisector of *OP*, which cuts *OP*
at *M*.

Step 4: With *M* as centre and *MO* as radius, draw a
circle which cuts previous circle at *A* and *B*.

Step5: Join *AP* and *BP*. *AP* and *BP* are
the required tangents. Thus length of the tangents are *PA* = *PB* =
7.4 cm.

**Verification : **In the right angle triangle** ***OAP***
**,** ***PA*^{2}** **=** ***OP*^{2}**
**−*OA*^{2}** **= 64 – 9 = 55

PA = √55 = 7 4. cm (approximately) .

Tags : Theorem, Statement, Proof, Construction, Solved Example Problems | Geometry Theorem, Statement, Proof, Construction, Solved Example Problems | Geometry

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