Chapter – 15 : THEOREMS RELATED TO TANGENT OF A CIRCLE

Question 1

But in no. (iii) in what position are the stick AB and the circular ring?

Solution:

The stick AB touches the circular ring at the point P.

Like me, my friend Sumedha similarly drew in what positions a straight line and a circular ring can possibly be to each other in her copy.

We see, in picture no. (i) straight line AB does not intersect the circle.

Again, in picture no. (ii) the straight line AB intersects the circle at the points P and Q.

Question 2

But in picture no. (ii) and (iii) what is this straight line AB of the circle called?

Solution:

In picture no. (ii) AB is a secant of the circle, and PQ is a corresponding chord of secant AB. We see that in picture no (iii) the common point of the circle and straight line AB is P.

∴ The straight line AB touches the circle at the point P.

3. But in picture no. (iii) what is the straight line AB of the circle called?

Solution:

In picture no. (iii) the straight line AB of the circle is called a tangent.

LET US WORK OUT – 15.1

Question 1

Masum has drawn a circle with center ‘O’, of which AB is a chord. I draw a tangent at the point B, which intersects the extended AO at the point T. If ∠BAT = 21°, let us write by calculating the value of ∠BTA.

Solution:

∠TOB = ∠BAT + ∠ABO = 2∠BAT = 42°.

Again in ∆ BOT, ∠OBT = 90°.

∴ ∠BTO = 180° – (∠TOB + ∠OBT) = 180° – (90° + 42°) = 48°.

Question 2

XY is a diameter of a circle. PAQ is a tangent to the circle at the point ‘A’ lying on the circumference. The perpendicular drawn on the tangent of the circle from

X intersects PAQ at Z. Let us prove that XY is a bisector of ∠YXZ.

Solution:

Let O be the center of the circle.

Join A, O as PAQ is a tangent.

∴ ∠OAZ = 90°.

Again, ∠OAZ = ∠OAX + ∠AZX.

∠OAX = ∠OXA [∆ OAX is an isosceles triangle].

∴ 90° = ∠OXA + ∠ZAX, and 90° = ∠ZAX + ∠ZXA.

Hence, ∠OXA + ∠ZAX = ∠ZAX + ∠ZXA

or, ∠OXA = ∠ZXA.

∴ XA is the bisector of ∠YXZ.

Question 3

I drew a circle having PR as a diameter. I draw a tangent at the point P and a point S is taken on the tangent of the circle in such a way that PR = PS. If RS intersects the circle at the point T, let us prove that ST = RT = PT.

Solution:

As ∠PTR is a semicircle angle.

∴ ∠PTR = ∠TSP + ∠TPS

Again, 90° = ∠TSP + ∠RPT

∴ ∠TSP = ∠RPT

(i) ∠TSP = ∠RPT (ii) PR = PS & (iii) PT is common.

∴ ∆PTS ≅ ∆PTR

Question 4

Two radii OA and OB of a circle with centre O are perpendicular to each other. If two tangents drawn at the point A and B intersect each other at the point T, let us prove that AB = OT and they bisect each other at a right – angle.

Solution:

Let O be the centre of the circle.

In quadrilateral AOBT, AO = OB

& ∠AOB = ∠TAO = ∠TBO = 90°

AOBT is a square.

∴ Diagonal AB = OT.

They bisect each other perpendicularly.

Question 5

Two chords AB and AC of the larger of two concentric circles touch the other circle at points P and Q respectively. Let us prove that PQ $= \frac{1}{2}$ BC.

Solution:

Let O is the centre of two concentric circles. AB & AC the two chords of the big circle, are the tangents of the small circles: Join A, O; B, O; C, O.

∴ AO = CO (Radius)

Again, OQ ⊥ AC.

∴ AQ = QC

Similarly, AP = PB.

∴ In ∆ABC, P & Q are the midpoints of AB & AC.

∴ PQ $= \frac{1}{2}$ BC

Question 6

X is a point on the tangent at the point A lies on a circle with centre O. A secant drawn from a point X intersects the circle at the points Y and Z If P is a mid-point of Y Z, let us prove that XAPO or XAOP is a cyclic quadrilateral.

Solution:

YZ is a chord of the circle. P is the midpoint of Y Z.

∴ OP ⊥ YZ

Again, XA is a tangent at A.

∴ AO ⊥ XY

∴ In the Quadrilateral XAPO,

∠OAX &, ∠XPO are equal = 90°

∴ XAPO is a cyclic quadrilateral whose XO is a diameter.

Question 7

P is any point on diameter of a circle with centre O. A perpendicular drawn on diameter at the point O intersects the circle at the point Q. Extended QP intersects the circle at the point R. A tangent drawn at the point R intersects extended OP at the point S. Let us prove that SP = SR.

Solution:

∠ORP + ∠PRS = 90°

or, ∠OQP + Q₁QP = 90°

∴ ∠ORP = ∠OQP [ as OQ = OR]

∴ ∠PRS = ∠Q₁QP ——————– (i)

∴ ∠OPQ = ∠SPR ———————— (ii) (Vertically opposite)

Again, ∠OPQ + ∠OQP = 90°

& ∠OQP + ∠PQQ₁ = 90°

∴ ∠OQ = ∠PQQ₁

∴ ∠SPR = ∠PRS

∴ ∆In SPR, SP = SR Proved.

Question 8

Rumela drew a circle with centre O, of which QR is a chord. Two tangents drawn at the points Q and R intersect each other at the point P. If QM is a diameter, let us prove that ∠QPR = 2 ∠RQM.

Solution:

In Quadrilateral PQOR, ∠PQO = ∠PRO = 90°

∴ Opposite angles of the Quadrilateral PQOR are supplementary.

∴ PQOR is a cyclic quadrilateral whose ∠ORP = 90°.

∴ PQOR is a square (OQ = OR).

∴ ∠ROQ = 90° = ∠QPR

& ∆In QOR, OQ = OR.

∴ ∠RQM = 45°

∴ ∠OPR – 2 ∠RQM. Proved.

Question 9

Two chords AC and BD of a circle intersect each other at the point O. If two tangents drawn at the points A and B intersect each other at the point P and two tangents drawn at the points C and D intersect at the point Q, let us prove that ∠P + ∠Q = ∠2 BOC.

Solution:

Let ‘ R ‘ is the centre.

∴ ARBP is a cyclic quadrilateral.

∴ ∠A + ∠B = 180° ∴ ∠ARB + ∠P = 180°

∴ ∠ARB = 2 ∠ACB

∴ 2 ∠ARB + ∠P = 180° ——————– (i)

Similarly, 2 ∠DAO + ∠Q = 180°

or, 2 ∠CBO + ∠Q = 180° ——————– (ii)

∴ 2 ∠ACB + 2 ∠CBO + ∠P + ∠Q = 360°

Adding (i) & (ii),

∠P + ∠Q = 2[180 – (∠ACB + ∠CBO)]

∴ ∠P + ∠Q = 2 ∠BOC.

(i) We take three coloured card board.

(ii) Drawing three circles with same measures with centre O on a white paper and cut off the circular region and pasted them on this coloured paper.

Solution:

We see that in picture no. (i), i.e., two tangents can be drawn to a circle from any point outside it.

In picutre no. (ii) i.e., one tangent can be drawn to a circle from a point lying on the circle.

We understand that, from any ____________________ [interior/external] point, no tangent can be drawn to the circle.

Ans. Interior.

Application 10.

If I draw a circle with centre O, a point P is 26 cm. away from the centre of the circle and the length of the tangent drawn from the point P to the circle is 10 cm, let us write by calculating the length of radius of the circle.

Solution:

PT = 10 cm., PO = 26 cm.

∴ From right-angle triangle POT, we get,

PO² = PT² + OT² [we get from Pythagoras theorem]

∴ (26 cm)² = (10 cm) 2 + PT²

or, OT² = (26 cm)² – (10 cm)² = 576 cm

∴ OT = 24 cm.

LET US DO – 21.1

Question 1

Let us prove that the line segment joining a point outside a circle and the centre bisects the angle included by two tangents drawn from the external point.

Solution:

O is the centre of the circle and two tangents PA & PB are drawn from the external point P.

Join O, A & O, B; O, P

In ∆AOP & ∆BOP

(i) OA = OB; OP common

& ∠OAP = ∠OBP common

∴ ∆OAP ≅ ∆OBP

∴ ∠APO = ∠BPO

Question 2

Let us prove that the bisector of angle included by two tangents to a circle from a point outside it passes through the centre.

Solution:

Let O is the centre of the circle and two tangents PA & PB are drawn from the external point P. Join O, A, O, B & A, B.

PD, the bisector of ∠P, cuts AB at D.

∠OAP = ∠OBP = 90°

or, ∠OAD + ∠DAP = ∠OBD + ∠DBP

∠DAP = ∠DBP

Now in ∆ABD & ∆BPD,

(i) ∠APD = ∠BPD, (ii) ∠DAP = ∠DBP

(iii) DP is common

∴ ∆APD ≅ ∆BPD

∴ AD – BD & ∠ADP = ∠BDP = 90°

Again, in ∆OAD & ∆OBD, (i) OA = OB; (ii) ∠OAD = ∠OBD ; (iii) AD = BD.

∴ ∆OAD ≅ ∆OBD

∴ ∠ODA = ∠ODB = 90°

∴ AB ⊥ OD & AB ⊥ PD

∴ Produced PD will pass through the centre. Proved.

Question 3

Let us prove that if two tangents drawn to a circle at two points on it intersect each other, then the lengths of the line segments from the point of intersection to the points of contact are equal.

Solution:

Let P & Q are two points on a circle with centre O. Two tangents drawn at P & Q intersect each other at A. To prove AP = AQ.

Joint O, P, O, Q & O, R

In ∆AOQ & ∆AOP,

(i) OP = OQ; OA common

&∠APO = ∠BQP = 90°

∴ ∆AOP ≅ ∆AOQ

∴ AP = AQ Proved.

Question 5

I see the position of circular rings as being put by Rabeya and let us see the different positions of rings.

Solution:

We see that in picture no. (i) two circles are concentric but in picture no. (ii) two circles are not concentric circles, in picture no. (iii) two circles do not intersect each other, in picture no. (iv) the two circles intersect each other at two points.

Question 7

How shall we understand that two circle touch each other?

Solution:

There is only one intersecting point P of two circles with centre O and O’ in the same plane and it the tangent of circle with centre O touches the circle with centre O’ at that point, then it is called that the two circles touch each other at the point P. In picture no. (vi) two circles do not touch each other. In picture no. (viii) and (ix) two circles touch each other. In picture no. (vii) two circles touch internally each other at the point P. AB is a common tangent ot two circles. Again, in picture no. (ix) two circles touch externally each other at the point P. AB is a common tangent of the two circles.

Question 9

But sometimes we see that the two circles are situated on the same side of the common tangents, again some other times the two circles are situated on the opposite sides of the common tangents. What is this type of tangent called?

Solution:

If two circles lie on same side of a common tangent the tangent, is said to be direct common tangent and if two circles lie on opposite side of common tangent, the common tangent is said to be transverse common tangent.

We understand that the tangents of picture no. (x) and (xi) are direct common tangents. But in picture no. (xii) there are two direct common tangents and 1 transverse common tangent. Again in picture no. (xiii) there are two direct common tangents and two transverse

Let us prove with reason that if two circles touch each other, the straight line through the centre of one circle and the point of constant passes through the centre of other circle. [Let me do it myself]

Solution:

Let two circles with centres C₁ & C₂ touched each other at A externally, PQ is a common tangent passing through A ; Join A, C₁ & A, C₂.

C₁A PQ & C₂ A PQ

∴ PAC₁ = PAC₂ = 90°

$\because$ C₁AC₂ = 180°

∴ C₁C₂ straight line, passing through A.

In the picture beside, PQ = AP + AQ.

Application 15.

In the adjoining figure, the incircle of ∆ABC touches the sides AB, BC and CA at the points D, E and F respectively. Let us prove that AD + BE + CF = AF + CE + BD = $\frac{1}{2}$ (perimeter of ∆ABC)

Solution:

Join A, O; B, O; C, O.

In ∆BOD & ∆BOE,

(i) ∠BDO = ∠BED

(ii) OD = OE (Radius)

(iii) OB is common

∴ ∆BOD ≅ ∆BOE, ∴ BD – BE ————— (i)

Similarly, from ∆COE & ∆COF, CE = CF —————— (ii)

Similarly, from ∆AOD & ∆AOF, AF = AD ———————- (iii)

Adding (i), (ii) & (iii),

AD + BE + CF = AF + CE + BD

Perimeter of ∆ABC

= AB + BC + CA

= AD + BD + BE + CE + CF + AF

= 2(AD + BE + CF)

∴ AD + BE + CF = AF + CE + BD $= \frac{1}{2}$ (AB + BC + CA)

$= \frac{1}{2}$ of perimeter of ∆ABC Proved.

LET US SEE BY CALCULATING – 15.2

Question 1

An external point is situated at a distance of 17 cm from the centre of a circle having 16 cm diameter, let us determine the length of the tangent drawn to the circle from the external point.

Solution:

Let the diameter of the circle = 16 cm

∴ Radius $= \frac{16cm} {2}$ = 8 cm

Distance of the external point from the centre (OP) = 17 cm

∴ In right-angled ∆OAP,

AP $= \sqrt{O P^2 - O A^2} = \sqrt{(17)^2 - (8)^2}$

$= \sqrt{289 - 64} = \sqrt{225 = 15}$

∴ AP = 15 cm

Question 2

The tangents drawn at points P and Q on the circumference of a circle intersect at A. If ∠PAQ = 60°, let us find the value of ∠APQ.

Solution:

Let AP & AQ are two tangents to the circle with centre O. From an external point A.

∴ AP = AQ

∆APQ is an isosceles triangle.

∴ ∠APQ = ∠AQP

If ∠PAQ = 60°

∴ ∠APQ $= \frac{180° - 60°}{2}$

= 60°

Question 3

AP and AQ are two tangents drawn from an external point A to a circle with centre O. P and Q are points of contact. If PR is a diameter, let us prove that OA || RQ.

Solution:

AP & AQ are two tangents to the circle with centre O, from an external point A.

∴ PA = QA & ∠AOP = ∠AOQ

Now in ∆POR & ∆QOR,

OP = OQ, ∠POR = ∠QOR & OR is common.

∴ ∆POR ≅ ∆QOR

PR = QR

∠PRO = ∠QRO

OR is standing on PQ

∠PRO = ∠QRO

∴ QR ⊥ PQ

i.e., OA ⊥ OQ

Again, PR = QR

∴ OR ⊥ RQ

Question 4

Let us prove that for a quadrilateral circumscribed about a circle, the angles subtended by any two opposite sides at the centres are supplementary to each other.

Solution:

‘ O ‘ is the centre of the circle. ABCD is a quadrilateral circumscribing the circle.

Let the circle touches the sides of the quadrilateral AB, BC, CD & DA at P, Q, R, S respectively.

∴ To prove, ∠AOB + ∠COD = 180° or ∠AOD + ∠BOC = 180°

Proof : As AP & AS are two tangents of the circle from A

∴ ∠AOP = ∠AOS

Similarly, ∠BOP = ∠BOQ

∴ ∠AOB = ∠AOP + ∠BOP = ∠AOS + ∠BOQ

Again, DS & DR are two tangents as S & R

∴ ∠DOS = ∠DOR

Similarly from tangents R & Q

∠COR = ∠COQ

∴ ∠COD = ∠COR + ∠DOR = ∠COQ + ∠DOS

∴ ∠AOB + ∠COD = ∠AOS + ∠BOQ + ∠COQ + ∠DOS

= (∠AOS + ∠DOS) + (∠BOQ + ∠COQ)

= ∠AOD + ∠BOC

Again, ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°

or, (∠AOB + ∠COD) + (∠AOB + ∠COD) = 360°

∴ 2 × (∠AOB + ∠COD) = 360°

∴ ∠AOB + ∠COD $= \frac{360°}{2}$ = 180°

or, 2(∠AOD + ∠BOC) = 360°

∴ ∠AOD + ∠BOC $= \frac{360°}{2}$ = 180° Proved.

Question 5

Let us prove that a parallelogram circumscribed by a circle is a rhombus.

Solution:

Let O is the centre of the circle & ABCD is a parallelogram circumscribing the circle. The circle touches the parallelogram at P, Q, R & S.

Join AO, PQ, BO, QO, RO & SO.

To prove A BCD is a rhombus

Proof : AP & AS are two tangents at P & S.

∴ AP = AS & ∠AOP = ∠AOS

Similarly, in case of tangent BP & BQ

BP = PQ ∠BOP = ∠BOQ

Again, in case of tangents SD & RD

SD = RD & ∠ODS = ∠ODR

Now in ∆s BOP & BOQ,

(i) ∠OPB = ∠OQB = 90°

(ii) ∠BOP = ∠BOQ & BP = BQ

∴ ∆BOP ≅ ∆BQO

∴ ∠PBO = ∠QBO

Similarly, from ∆DOS & ∆DOR we get ∠SDO = ∠RDO

Now, in parallogram ABCD, ∠ABC & ∠ADC are two opposite angles.

∴ ∠ABC = ∠ADC ∴ ∠PBO = ∠RDO

Now in ∆s BOP & ROD

∠BPO = ∠DRO ; ∠PBO = ∠RDO & OP = OR

∴ ∆BOP ≅ ∆ROD

∴ PB = DR

∴ PB = SD

Now, AP + PB = AS + SD

i.e., AB = AD

∴ AB = BC = CD = DA

∴ All the sides of the parallelogram ABCD are equal.

∴ ABCD is a rhombus.

Question 6

Two circles drawn with centres A and B touch each other externally at C. O is a point on the tangent drawn at C, OD and OE are tangents drawn to the two circles of centres A and B respectively. If ∠COD = 56°, ∠COE = 40°, ACD = x° and ∠BCE = y°, let us prove that OD = OC = OE and y – x = 8

Solution:

∠COD = 56°, ∠COE = 40°, ∠ACD = x° & ∠BCE = y°

To prove OD = OC = OE & y – x = 8

Proof : OD & OC are two tangents to the circle with centre A at the points D and C.

∴ OD = OC & ∠ODC = ∠OCD

Again, OD & OE are two tangents of the circle with centre B at the points C & E

∴ OC = OE & ∠OEC = ∠OCE

∴ OD = OC = OE

∠COD = 56° & ∠COE = 40° (given)

Let ∠ACD = x° & ∠BCE = y°

∴ ∠OCD = \frac{180° – 56°}{2} = \frac{124°}{2} = 62°

OC is the common tangents of the two circles & AB is straight line joining the centres.

∴ OC ⊥ AB

∴ ∠OCA = ∠OCB = 90°

∠ACD = 90° – ∠O C D = 90° – 62° = 28° ∴ x = 28°

Again, ∠OCE $= \frac{180° - 40°}{2} = \frac{140°}{2} = 70^{\circ}$ ∴ ∠BCE

= 90° – ∠OCE = 90° – 70° = 20°

∴ y = 20° ∴ x – y = 28° – 20° = 8 Proved.

Question 7

Two circles with centres A and B. touch each other internally. Another circle touches the larger circle externally at the point x and the smaller circle externally at the point y. If O be the centre of that circle, let us prove that AO + BO is constant.

Solution:

To nrove AO + BO = Constant Join O, X, O, Y, O, A & Y, B

Two circles with centres A & B

The point of contact is on the line joining centre A & centre B.

Again, the two circles with centre O & centre B, touch externally at Y.

The point Y is on the line joining YB.

Again, the two circles with centre O & A touch internally at X.

∴ The point X is on the line joining OX.

Now, OA = AS – OX & OB = OY + YB

∴ AO + BO = AX – OX + OY + YB = AX – OX + OX + YB

AO + BO = AX + YB

= Constant Proved. [as OX = OY = radius of the circle with centre O & AX = Radius of the circle with centre A ; YB = Radius of the circle with centre B.]

Question 8

Two circles have been drawn with centres A and B which touch each other externally at the point O. I draw a straight line passing through the point O which intersects the two circles at P and Q respectively. Let us prove that AP || BQ.

Solution:

To prove AP || BQ

Join A, P & B, Q

In ∆AOP, A O = AP (Radius)

∴ ∆AOP is an issoceles triangle.

∴ ∠APO = ∠AOP

In ∆BOQ, BO = BQ (Radius)

∴ ∆B O Q is an issoceles triangle.

∴ ∠BOQ = ∠QOB

Now, the two circles with centres A & B touch externally at O, A B is the line joining their centres.

∴ ∠AOP = ∠BOQ (vertically opposite)

∴ ∠APO = ∠BQO

PQ cuts AP & BQ at P & Q, & produces two equal angles.

∴ They are alternate angles.

∴ AP || BQ Proved.

Question 9

Three equal circles touch one another externally. Let us prove that the centres of the three circles form an equilateral triangle.

Solution:

Let the circles with centres A, B, C touch externally at points P, Q and R. To prove ABC is an equilateral triangle.

Proof : Two circles with centres A & B touch each other at P.

∴ The point P is on the line, AB, joining the centre.

Similarly, the point Q is on the line BC & the point R is on the line CA.

As the radii of the three circles are equal, AP = PB = BQ = QC = CR + RA

∴ AP + PB = BQ + QC = CR + RA

i.e., AB = BC = CA

∴ ∆ABC is an equilateral triangle.

Question 10

Two tangents AB and AC drawn from an external point A of a circle touch the circle at the points B and C. A tangent drawn to a point X lies on the minor arc BC intersects AB and AC at the points D and E respectively. Let us prove that which perimeter of ∆ADE = 2 AB.

Solution:

To prove the perimeter of ∆ADE = 2 AB

Let O is the centre of the circle.

Join O, B, O, C, O, D & O, X.

In Quadrilateral OXDB

∠OBD = ∠OXD = 90°

∴ OXDB is a cyclic quadrilateral whose OD is a diagonal.

Now in ∆OBD & ∆OXD,

(i) OB = OX (Radii of same circle)

(ii) OD common

& (iii) ∠OBD = ∠OXD = 90°

∴ ∆OBD ≅ ∆OXD

∴ BD = DX

Similarly CE = XE

∴ (AD + DX) + (AE + EX) = (AD + BD) + (AE + CE)

= AB + AC = 2 AB[ $\because$ AB = AC]

Again, (AD + DX) + (AE + EX) AD + DE + AE = Perimeter of ∆ADE

∴ Perimeter of ∆ADE = 2 AB Proved.

11. Very short answer type. (V.S.A.)

(A) M.C.Q.

Question 1

A tangent drawn to a circle with centre O from an external point A touches the circle at the point B. If OB = 5 cm, AO = 13 cm, then the length of AB is

12 cm
13 cm
6.5 cm
6 cm

Solution : AB² = AO² – OB² = 13² – 5² = 169 – 25 = 144

∴ AB = 12 cm.

Solution:

(a) 12 CM

Question 2

Two circles touch each other externally at the point C. A direct common tangent A B touches the two circles at the points A and B. The value of ∠ACB is

60°
45°
30°
90°

Solution:

(d) 90°

Question 3

The length of radius of a circle with centre O is 5 cm. P is a point at the distance of 13 cm from the point O. The length of two tangents are PO and PR from the point P. The area of quadrilateral PQOR is

60 sq cm.
30 sq cm.
120 sq cm.
150 sq cm.

Solution:

Area of quadrilateral

= ∆PQR + ∆POQ

= ( $\frac{1}{2}$ × 5 × 12) + ( $\frac{1}{2}$ × 5 × 12) sqm

= 60 sq cm.

Ans. (a) 60 sq cm.

Question 4

The lengths of radii of two circles are 5 cm and 3 cm. The two circles touch each other externally. The distance between two centres of the two circles is

2 cm.
2.5 cm.
1.5 cm.
8 cm.

Solution:

The distance between the centres

= Sum of their radii

= 5 cm + 3 cm = 8 cm

Ans. (d) 8 cm.

Question 5

The lengths of radii of two circles are 3.5 cm and 2 cm. The two circles touch each other internally. The distance between the centres of the two circles is

5.5 cm.
1 cm.
1.5 cm.
None of these

Solution:

The distance between the centres

= Difference between their radii

(3.5 – 2) cm = 1.5 cm

Ans. (c) 1.5 cm.

(B) Let us write whether the following statements are true or false :

Question 1

P is a point inside a circle. Any tangent drawn on the circle does not pass through the point P.

Solution:

TRUE

Question 2

There are more than two tangents which can be drawn to a circle parallel to a fixed Line.

Solution:

FALSE

(C) Let us fill in the blanks.

Question 1

If a straight line intersects the circles at two points, then the straight line is ________________ of circle.

Solution:

Bisector.

Question 2

If two circles do not intersect or touch each other, then the maximum number of common tangents that can be drawn is _______________.

Solution:

4. (Four).

Question 3

Two circles touch each other externally at the point A. A common tangent drawn to the two circles at the point A is ____________ common tangents (direct/transverse)

Solution:

Transverse.

Short answer type questions (S.A.)

Question 1

In the adjoining figure O is the centre and BOA is a diameter of the circle. A tangent drawn to a circle at the point P intersects the extended BA at the point T. If ∠PBO = 30°, let us find the value of ∠PTA.