How many tangents can a circle have ?

SOLUTION:
**Soln. :** A circle can have an infinite number of tangents.

Fill in the blanks :

(i) A tangent to a circle intersects it in .......... point(s).

(ii) A line intersecting a circle in two points is called a ........... .

(iii) A circle can have .......... parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called ......... .

**Soln. :** (i) exactly one (ii) secant

(iii) two (iv) point of contact

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* = 12 cm. Length *PQ* is

(A) 12 cm

(B) 13 cm

(C) 8.5 cm

(D) cm

**Soln. :**(D) : In right ∆*POQ*,
*OQ*^{2} = *OP*^{2} + *PQ*^{2}

⇒

Draw a circle and two lines parallel to a given line such that one is a tangent and the other a secant to the circle.

SOLUTION:
**Soln. :** We have the required figure, as shown

Here, *l* is the given line and a circle with centre *O* is drawn.

The line *n* is drawn which is parallel to *l* and tangent to the circle. Also, *m* is drawn parallel to line *l* and is a secant to the circle.

Choose the correct option:

From a point *Q*, the length of the tangent to a circle is 24 cm and the distance of *Q* from the centre is 25 cm. The radius of the circle is

(A) 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

**Soln. :** (A) ∵ *QT* is a tangent to the circle at *T* and *OT* is radius

∴ *OT* ┴ *QT*

Also, *OQ* = 25 cm

and *QT* = 24 cm

∴ Using Pythagoras theorem, we get
*OQ*^{2} = *QT*^{2} + *OT*^{2}

⇒ *OT*^{2} = *OQ*^{2} - *QT*^{2} = 25^{2} - 24^{2} = 49

⇒ *OT* = 7

Thus, the required radius is 7 cm.

In figure, if *TP* and *TQ* are the two tangents to a circle with centre *O* so that ∠*POQ* = 110°, then ∠*PTQ* is equal to

(A) 60°

(B) 70°

(C) 80°

(D) 90°

**Soln. :** (B) ∴ *TQ* and *TP* are tangents to a circle with centre *O* and ∠*POQ* = 110°

∴ *OP* ┴ *PT* and *OQ* ┴ *QT*

⇒ ∠*OPT* = 90° and ∠*OQT* = 90°

Now, in the quadrilateral *TPOQ*, we get

∴ ∠*PTQ* + 90° + 110° + 90° = 360°

[Angle sum property of a quadrilateral]

⇒ ∠*PTQ* + 290° = 360°

⇒ ∠*PTQ* = 360° - 290° = 70°

If tangents *PA* and *PB* from a point *P* to a circle with centre *O* are inclined to each other at angle of 80°, then ∠*POA* is equal to

(A) 50°

(B) 60°

(C) 70°

(D) 80°

**Soln. :** (A) : Since, *O* is the centre of the circle and two tangents from *P* to the circle are *PA* and *PB*.

∴ *OA* ┴ *AP* and *OB* ┴ *BP*

⇒ ∠*OAP* = ∠*OBP* = 90°

Now, in quadrilateral *PAOB*, we have :

∠*APB* + ∠*PAO* + ∠*AOB* + ∠*PBO* = 360°

⇒ 80° + 90° + ∠*AOB* + 90° = 360°

⇒ 260° + ∠*AOB* = 360°

⇒ ∠*AOB* = 360° - 260° ⇒ ∠*AOB* = 100°

In right ∆*OAP* and right ∆*OBP*, we have
*OP* = *OP* [Common]

∠*OAP* = ∠*OBP* [Each = 90°]
*OA* = *OB* [Radii of the same circle]

⇒ ∠*POA* = ∠*POB* [By CPCT]

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

SOLUTION:
**Soln. :** In the figure, *PQ* is diameter of the given circle and *O* is its centre.

Let tangents *AB* and *CD* be drawn at the end points of the diameter *PQ*.

Since the tangents at a point to a circle is perpen-dicular to the radius through the point.

∴ *PQ* ┴ *AB*

⇒ ∠*APQ* = 90°

And *PQ* ┴ *CD*

⇒ ∠*PQD* = 90°

⇒ ∠*APQ* = ∠*PQD *

But they form a pair of alternate angles.

∴ *AB* || *CD*.

Hence the two tangents are parallel.

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

SOLUTION:
**Soln. :** In the figure, the centre of the circle is *O* and tangent *AB* touches the circle at *P*. If possible, let *PQ* be perpendicular to *AB* such that it is not passing through *O*.

Join *OP*.

Since tangent at a point to a circle is perpendicular to the radius through that point,

∴ *OP* ┴ *AB*

⇒ ∠*OPB* = 90° ...(1)

But by construction,
*PQ* ┴ *AB*

⇒ ∠*QPB* = 90° ...(2)

From (1) and (2),

∠*QPB* = ∠*OPB *

which is possible only when *O* and *Q* coincide. Thus, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

The length of a tangent from a point *A* at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

**Soln. :** ∵ The tangent to a circle is perpendicular to the radius through the point of contact.

∴ ∠*OTA* = 90°

Now, in the right ∆*OTA*, we have:
*OA*^{2} = *OT*^{2} + *AT*^{2} [Pythagoras theorem]

⇒ 5^{2} = *OT*^{2} + 4^{2}

⇒ *OT*^{2} = 5^{2} - 4^{2}

⇒ *OT*^{2} = (5 - 4)(5 + 4)

⇒ *OT*^{2} = 1 × 9

= 9 = 3^{2}

⇒ *OT* = 3

Thus, the radius of the circle is 3 cm.

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

SOLUTION:
**Soln. :** In the figure, *O* is the common centre, of the given concentric circles.
*AB* is a chord of the bigger circle such that it is a tangent to the smaller circle at *P*.

Since *OP* is the radius of the smaller circle. ∴ *OP* ┴ *AB* ⇒ ∠*APO* = 90°

Also, radius perpendicular to a chord bisects the chord.

∴ *OP* bisects *AB*

⇒

Now, in right ∆*APO*,
*OA*^{2} = *AP*^{2} + *OP*^{2}

⇒ 5^{2} = *AP*^{2} + 3^{2} ⇒ *AP*^{2} = 5^{2} - 3^{2}

⇒ *AP*^{2} = 4^{2} ⇒ *AP* = 4 cm

⇒

Hence, the required length of the chord *AB* is 8 cm.

A quadrilateral *ABCD* is drawn to circumscribe a circle (see figure).

Prove that *AB* + *CD* = *AD* + *BC*

**Soln. :** Since the sides of quadrilateral *ABCD*, *i.e*., *AB, BC, CD* and *DA* touch the circle at *P, Q, R* and *S* respectively, and the lengths of two tangents to a circle from an external point are equal.

∴ *AP* = *AS*
*BP* = *BQ*
*DR* = *DS* and *CR* = *CQ*

Adding them, we get

(*AP* + *BP*) + (*CR* + *RD*) = (*BQ* + *QC*) + (*DS* + *SA*)

⇒ *AP* + *CD* = *BC* + *DA*

In the figure, *XY* and *X*'*Y*' are two parallel tangents to a circle with centre *O* and another tangent *AB* with point of contact *C* intersecting *XY* at *A* and *X*'*Y*' at *B*.

Prove that ∠*AOB* = 90°.

**Soln. :** ∵ The tangents drawn to a circle from an external point are equal.

∴ *AP* = *AC*, Join *OC*

In ∆*PAO* and ∆*CAO*, we have :
*AO* = *AO* [Common]
*OP* = *OC* [Radii of the same circle]
*AP* = *AC* [Proved above]

⇒

∴ ∠*PAO* = 2∠*CAO*

⇒ ∠*PAC* = 2∠*CAO* ...(1)

Similarly ∠*CBQ* = 2∠*CBO* ...(2)

Again, we know that sum of internal angles on the same side of a transversal is 180°.

∴ ∠*PAC* + ∠*CBQ* = 180°

⇒ 2∠*CAO* + 2∠*CBO* = 180°

[From (1) and (2)]

⇒ ...(3)

Also in ∆*AOB*, ∠*BAO* + ∠*ABO* + ∠*AOB* = 180°

[Sum of angles of a triangle]

⇒ ∠*CAO* + ∠*CBO* + ∠*AOB* = 180° [By (3)]

⇒ 90° + ∠*AOB* = 180°

⇒ ∠*AOB* = 180° - 90°

⇒ ∠*AOB* = 90°.

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

SOLUTION:
**Soln. :** Let *PA* and *PB* be two tangents drawn from an external point *P* to a circle with centre *O*.

Now, in right ∆*OAP* and right ∆*OB*P, we have
*PA* = *PB* [Tangents to circle from an

external point]
*OA* = *OB* [Radii of the same circle]
*OP* = *OP* [Common]

[By SSS congruency]

∴ ∠*OPA* = ∠*OPB* [By C.P.C.T.]

and ∠*AOP* = ∠*BOP*

⇒ ∠*APB* = 2∠*OPA* and ∠*AOB* = 2∠*AOP*

But ∠*AOP* = 90° - ∠*OPA*

⇒ 2∠*AOP* = 180° - 2∠*OPA*

⇒ ∠*AOB* = 180° - ∠*APB*

⇒ ∠*AOB* + ∠*APB* = 180°. (Proved)

Prove that the parallelogram circumscribing a circle is a rhombus.

SOLUTION:
**Soln. :** We have *ABCD*, a parallelogram which circumscribes a circle (*i.e*., its sides touch the circle) with centre O.

Since tangents to a circle from an external point are equal in length,

∴ *AP* = *AS*
*BP* = *BQ*
*CR* = *CQ*
*DR* = *DS*

Adding, we get

(*AP* + *BP*) + (*CR* + *DR*)

= (*AS* + *DS*) + (*BQ* + *CQ*)

⇒ *AB* + *CD* = *AD* + *BC*

But *AB* = *CD* [opposite sides of parallelogram]

and *BC* = *AD*

∴ *AB* + *CD* = *AD* + *BC* ⇒ 2*AB* = 2*BC*

⇒ *AB* = *BC*

Similarly *AB* = *DA* and *DA* = *CD*

Thus, *AB* = *BC* = *CD* = *AD*

Hence *ABCD* is a rhombus.

A triangle *ABC* is drawn to circumscribe a circle of radius 4 cm such that the segments *BD* and *DC* into which *BC* is divided by the point of contact *D* are of lengths 8 cm and 6 cm respectively (see figure). Find the sides *AB* and *AC*.

**Soln. :** Here ∆*ABC* circumscribe the circle with centre *O*. Also, radius = 4 cm

∵ The sides *BC*, *CA* and *AB* touch the circle at *D*, *E* and *F* respectively.

∴ *BF* = *BD* = 8 cm
*CD* = *CE* = 6 cm
*AF* = *AE* = *x* cm (say)

⇒ The sides of the triangle are :

14 cm, (*x* + 6) cm and (*x* + 8) cm Perimeter of ∆*ABC*

= [14 + (*x* + 6) + (*x* + 8)] cm

= [14 + 6 + 8 + 2*x*] cm

= 28 + 2*x* cm

⇒ Semi perimeter of ∆*ABC*

where *a* = *AB*, *b* = *BC*, *c* = *AC*

∴ *s* - *a* = (14 + *x*) - (8 + *x*) = 6
*s* - *b* = (14 + *x*) - (14) = *x*
*s* - *c* = (14 + *x*) - (16 + *x*) = 8

∴ Area of ∆*ABC* = cm^{2}

= ...(1)

Now, ar(∆*OBC*) =

= 28 cm^{2} [∵ *OD* = Radius]

ar (∆*OCA*)

= (2*x* + 12) cm^{2}

ar (∆*OAB*)

= (2*x* + 16) cm^{2}

∴ ar (∆*ABC*) = ar (∆*OBC*) + ar (∆*OCA*) + ar (∆*OAB*)

= 28 cm^{2} + (2*x* + 12) cm^{2} + (2*x* + 16) cm^{2}

= (28 + 12 + 16) + 4*x* cm^{2}

= (56 + 4*x*) cm^{2} ...(2)

From (1) and (2), we have

56 + 4*x* =

⇒ 4[14 + *x*] =

⇒ 14 + *x* =

Squaring both sides (14 + *x*)^{2} = (14 + *x*)3*x*

⇒ 196 + *x*^{2} + 28*x* = 42*x* + 3*x*^{2}

⇒ 2*x*^{2} + 14*x* - 196 = 0

⇒ *x*^{2} + 7*x* - 98 = 0

⇒ (*x* - 7) (*x* + 14) = 0

⇒ *x* - 7 = 0

⇒ *x* = 7 or *x* + 14 = 0

⇒ *x* = -14

But *x* = -14 is rejected.

∴ *x* = 7 cm

Thus, *AB* = 8 + 7 = 15 cm
*BC* = 8 + 6 = 14 cm
*CA* = 6 + 7 = 13 cm.

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

SOLUTION:
**Soln. :** We have a circle with centre *O*. A quadrilateral *ABCD* is such that the sides *AB*, *BC*, *CD* and *DA* touch the circle at *P*, *Q*, *R* and *S* respectively.

Let us join *OP*, *OQ*, *OR* and *OS*.

We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠2

∠3 = ∠4

∠5 = ∠6

and ∠7 = ∠8

Also, the sum of all the angles around a point is 360°.

∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8

= 360°

∴ 2[∠1 + ∠8 + ∠5 + ∠4] = 360°

⇒ (∠1 + ∠8 + ∠5 + ∠4) = 180° ...(1)

And 2[∠2 + ∠3 + ∠6 + ∠7] = 360°

⇒ (∠2 + ∠3) + (∠6 + ∠7) = 180° ...(2)

Since, ∠2 + ∠3 = ∠*AOB*, ∠6 + ∠7 = ∠*COD*

∠1 + ∠8 = ∠*AOD*, ∠4 + ∠5 = ∠*BOC*

∴ From (1) and (2), we have:

∠*AOD* + ∠*BOC* = 180°

and ∠*AOB* + ∠*COD* = 180°(Proved)