Number of Tangents from a Point on a Circle Questions in English

(A) circle is a collection of infinite points on a plane that are at a constant distance from a fixed point called the center. Since a tangent to a circle touches the circle at exactly one point,and there are infinite points on the circumference of a circle,there can be infinite tangents to a circle.

(C) Given that $PA$ and $PB$ are tangents to the circle from an external point $P$.
The radius drawn to the point of contact is perpendicular to the tangent.
Therefore,$OA \perp PA$ and $OB \perp PB$,which implies $\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.
In the quadrilateral $OAPB$,the sum of interior angles is $360^{\circ}$.
$\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^{\circ}$
$90^{\circ} + 80^{\circ} + 90^{\circ} + \angle AOB = 360^{\circ}$
$260^{\circ} + \angle AOB = 360^{\circ}$
$\angle AOB = 100^{\circ}$
In $\triangle OPA$ and $\triangle OPB$:
$OA = OB$ (Radii of the same circle)
$PA = PB$ (Tangents from an external point are equal)
$OP = OP$ (Common side)
By $SSS$ congruence criterion,$\triangle OPA \cong \triangle OPB$.
By $CPCT$,$\angle POA = \angle POB$.
Since $\angle POA + \angle POB = \angle AOB = 100^{\circ}$,we have $2 \angle POA = 100^{\circ}$.
Therefore,$\angle POA = 50^{\circ}$.

(A) Let the center of the circle be $O$ and the two radii be $OA$ and $OB$. The angle between the radii is $\angle AOB = 130^{\circ}$.
Let the tangents at points $A$ and $B$ meet at point $P$.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,$\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.
In the quadrilateral $OAPB$,the sum of the interior angles is $360^{\circ}$.
$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^{\circ}$
$130^{\circ} + 90^{\circ} + \angle APB + 90^{\circ} = 360^{\circ}$
$310^{\circ} + \angle APB = 360^{\circ}$
$\angle APB = 360^{\circ} - 310^{\circ} = 50^{\circ}$.
Thus,the angle between the tangents is $50^{\circ}$.

(B) Given that $AP$ and $AQ$ are tangents from an external point $A$ to a circle with centre $O$.
Since the tangents are perpendicular to each other,$\angle PAQ = 90^{\circ}$.
Join $OP$ and $OQ$. Also,join $OA$.
In quadrilateral $APOQ$,the radius is perpendicular to the tangent at the point of contact,so $\angle OPA = 90^{\circ}$ and $\angle OQA = 90^{\circ}$.
Since $\angle PAQ = 90^{\circ}$,$\angle OPA = 90^{\circ}$,and $\angle OQA = 90^{\circ}$,the fourth angle $\angle POQ$ must also be $90^{\circ}$ (as the sum of angles in a quadrilateral is $360^{\circ}$).
Thus,$APOQ$ is a square because $AP = AQ$ (tangents from an external point) and all its angles are $90^{\circ}$.
Therefore,the radius $OP = AP = 5 \, cm$.

(C) Given that $PA$ and $PB$ are tangents to the circle from an external point $P$.
Since the lengths of tangents drawn from an external point to a circle are equal,we have $PA = PB$.
In $\triangle PAB$,since $PA = PB$,the angles opposite to these sides are equal,i.e.,$\angle PBA = \angle PAB$.
Let $\angle PBA = \angle PAB = \theta$.
In $\triangle PAB$,the sum of angles is $180^{\circ}$,so $\angle APB + \angle PAB + \angle PBA = 180^{\circ}$.
Substituting the given values: $50^{\circ} + \theta + \theta = 180^{\circ}$.
$2\theta = 180^{\circ} - 50^{\circ} = 130^{\circ}$.
$\theta = 65^{\circ}$.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore,$OA \perp PA$,which means $\angle OAP = 90^{\circ}$.
Since $\angle OAP = \angle OAB + \angle PAB$,we have $90^{\circ} = \angle OAB + 65^{\circ}$.
$\angle OAB = 90^{\circ} - 65^{\circ} = 25^{\circ}$.

(D) Let $P$ be an external point from which two tangents $PA$ and $PC$ are drawn to a circle with center $O$ and radius $OA = 3 \, cm$.
The angle between the two tangents is $\angle APC = 60^{\circ}$.
Join $OP$. $OP$ is the angle bisector of $\angle APC$.
Therefore,$\angle APO = \angle CPO = \frac{60^{\circ}}{2} = 30^{\circ}$.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OA \perp AP$.
In the right-angled triangle $\triangle OAP$,we have:
$\tan(\angle APO) = \frac{OA}{AP}$
$\tan 30^{\circ} = \frac{3}{AP}$
$\frac{1}{\sqrt{3}} = \frac{3}{AP}$
$AP = 3 \sqrt{3} \, cm$.
Thus,the length of each tangent is $3 \sqrt{3} \, cm$.

(B) False
Given that a chord $AB$ subtends an angle of $60^{\circ}$ at the centre $O$ of a circle.
i.e.,$\angle AOB = 60^{\circ}$.
Since $OA = OB$ (radii of the same circle),$\triangle OAB$ is an isosceles triangle.
Therefore,$\angle OAB = \angle OBA = (180^{\circ} - 60^{\circ}) / 2 = 60^{\circ}$.
Let the tangents at points $A$ and $B$ intersect at point $C$.
We know that the radius is perpendicular to the tangent at the point of contact.
So,$OA \perp AC$ and $OB \perp BC$.
Therefore,$\angle OAC = 90^{\circ}$ and $\angle OBC = 90^{\circ}$.
Now,$\angle BAC = \angle OAC - \angle OAB = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Similarly,$\angle ABC = \angle OBC - \angle OBA = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
In $\triangle ABC$,the sum of interior angles is $180^{\circ}$.
So,$\angle ACB + \angle BAC + \angle ABC = 180^{\circ}$.
$\angle ACB + 30^{\circ} + 30^{\circ} = 180^{\circ}$.
$\angle ACB = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Thus,the angle between the tangents is $120^{\circ}$,not $60^{\circ}$.

(A) True.
The angle between two tangents to a circle can be $0^{\circ}$ if the two tangents are parallel to each other. In geometry,parallel lines are considered to have an angle of $0^{\circ}$ between them as they never intersect.

(A) True
Let the two tangents from point $P$ touch the circle at points $T$ and $R$. Given,the radius $OT = a$.
The line segment $OP$ bisects the angle between the two tangents,$\angle T P R$.
Therefore,$\angle T P O = \angle R P O = \frac{90^{\circ}}{2} = 45^{\circ}$.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OT \perp PT$.
In the right-angled triangle $\triangle OTP$,we have:
$\sin 45^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OT}{OP}$
Substituting the values,we get:
$\frac{1}{\sqrt{2}} = \frac{a}{OP}$
Therefore,$OP = a\sqrt{2}$.

(B) False.
Let $PT$ and $PR$ be the two tangents drawn from point $P$ to the circle with center $O$ and radius $a = OT = OR$.
The line $OP$ bisects the angle between the tangents,so $\angle TPO = \angle RPO = \frac{60^{\circ}}{2} = 30^{\circ}$.
Since the radius is perpendicular to the tangent at the point of contact,$\angle OTP = 90^{\circ}$.
In the right-angled triangle $\Delta OTP$,we have:
$\sin(\angle TPO) = \frac{OT}{OP}$
$\sin(30^{\circ}) = \frac{a}{OP}$
$\frac{1}{2} = \frac{a}{OP}$
$OP = 2a$.
Therefore,the statement $OP = a\sqrt{3}$ is False.

(N/A) Let the incircle of the right triangle $ABC$ touch the sides $BC, CA, AB$ at $D, E, F$ respectively,where $BC = a, CA = b$ and $AB = c$. Let $O$ be the center of the circle and $r$ be its radius.
Since the tangents drawn from an external point to a circle are equal in length,we have $AE = AF$,$BD = BF$,and $CD = CE = r$.
In the quadrilateral $CDOE$,since $\angle C = 90^\circ$ and the radii are perpendicular to the tangents,$\angle ODC = \angle OEC = 90^\circ$. Thus,$CDOE$ is a square with side $r$.
Therefore,$CD = CE = r$.
From the figure,$AF = AE = b - r$ and $BF = BD = a - r$.
Since $AB = c$,we have $AF + BF = c$.
Substituting the values,$(b - r) + (a - r) = c$.
$a + b - 2r = c$.
$2r = a + b - c$.
$r = \frac{a + b - c}{2}$.

(N/A) Two tangents $BD$ and $BC$ are drawn from an external point $B$.
To prove: $BO = 2BC$.
Join $OC$,$OD$,and $BO$.
Since $BC$ and $BD$ are tangents to the circle from an external point $B$,the line segment $BO$ bisects the angle $\angle DBC$.
Therefore,$\angle OBC = \angle DBO = \frac{1}{2} \angle DBC = \frac{1}{2} \times 120^{\circ} = 60^{\circ}$.
We know that the radius is perpendicular to the tangent at the point of contact. Thus,$OC \perp BC$ and $OD \perp BD$.
In the right-angled $\triangle OBC$,we have:
$\cos(\angle OBC) = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{BO}$
$\cos(60^{\circ}) = \frac{BC}{BO}$
$\frac{1}{2} = \frac{BC}{BO}$
$BO = 2BC$.
Since tangents drawn from an external point to a circle are equal in length,$BC = BD$.
Substituting $BD$ for $BC$,we get:
$BO = BC + BD$.

(N/A) Given: Two tangents $PR$ and $PQ$ are drawn from an external point $P$ to a circle with centre $O$.
To prove: The centre $O$ of the circle lies on the angle bisector of the angle formed by the two intersecting lines $PR$ and $PQ$.
Construction: Join $OR$ and $OQ$.
Proof:
In $\triangle PRO$ and $\triangle PQO$:
$1$. $\angle PRO = \angle PQO = 90^{\circ}$ (The tangent at any point of a circle is perpendicular to the radius through the point of contact).
$2$. $OR = OQ$ (Radii of the same circle).
$3$. $OP = OP$ (Common side).
By $R.H.S.$ congruence criterion,$\triangle PRO \cong \triangle PQO$.
Therefore,$\angle RPO = \angle QPO$ (by $CPCT$).
Since $\angle RPO = \angle QPO$,the line $OP$ bisects $\angle RPQ$. Thus,the centre $O$ lies on the angle bisector of the lines $PR$ and $PQ$. Hence proved.

(N/A) Given: Common tangents $AB$ and $CD$ to two circles intersect at $E$.
To prove: $AB = CD$.
Proof:
From point $E$,$EA$ and $EC$ are tangents to the left circle. Since the lengths of tangents drawn from an external point to a circle are equal,we have:
$EA = EC$ ......$(i)$
Similarly,from point $E$,$EB$ and $ED$ are tangents to the right circle. Therefore:
$EB = ED$ .......$(ii)$
Adding equations $(i)$ and $(ii)$,we get:
$EA + EB = EC + ED$
Since $EA + EB = AB$ and $EC + ED = CD$,we have:
$AB = CD$
Hence proved.

(B) False.
The radius of the circle is $r = 3.5 \, cm$ and the distance of point $P$ from the centre is $d = 3 \, cm$.
Since $d < r$ $(3 \, cm < 3.5 \, cm)$,the point $P$ lies inside the circle.
$A$ tangent to a circle is a line that touches the circle at exactly one point. If a point lies inside the circle,no tangent can be drawn from that point to the circle.

(N/A) circle is inscribed in the $\triangle ABC$,which touches the sides $BC, CA$ and $AB$ at $D, E$ and $F$ respectively.
Given,$BC = a, CA = b$ and $AB = c$.
By using the property that tangents drawn from an external point to a circle are equal in length:
$\therefore BD = BF = x$ (let)
$DC = CE = y$ (let)
$AE = AF = z$ (let)
The perimeter of $\triangle ABC = BC + CA + AB = a + b + c$.
Since $2s = a + b + c$,we have:
$(BD + DC) + (CE + EA) + (AF + FB) = a + b + c$
$(x + y) + (y + z) + (z + x) = a + b + c$
$2(x + y + z) = 2s$
$x + y + z = s$
We need to find $BD = x$.
From the equation $x + y + z = s$,we get $x = s - (y + z)$.
Since $b = CA = CE + EA = y + z$,substituting this into the equation:
$x = s - b$
Therefore,$BD = s - b$.
Hence proved.

(B) Given that $PA$ and $PB$ are tangents from an external point $P$ to a circle with centre $O$.
We know that the lengths of tangents drawn from an external point to a circle are equal,so $PA = PB = 10 \, cm$.
The perimeter of $\triangle PCD = PC + CD + PD$.
Since $CD$ is a tangent at point $E$,it can be written as $CD = CE + ED$.
Also,$CE = CA$ and $ED = DB$ (tangents from points $C$ and $D$ respectively).
Substituting these into the perimeter formula:
Perimeter of $\triangle PCD = PC + (CE + ED) + PD = PC + CA + DB + PD$.
Since $PC + CA = PA$ and $PD + DB = PB$,we get:
Perimeter of $\triangle PCD = PA + PB = 10 + 10 = 20 \, cm$.

(A) Given: Two tangents are drawn from an external point $A$ to the circle with centre $O$.
$OA = 13\, cm$,radius $OP = OQ = 5\, cm$.
Tangent $BC$ is drawn at a point $R$ on the minor arc $PQ$.
To find: Perimeter of $\triangle ABC$.
Proof: $\angle OPA = 90^{\circ}$ [Tangent at any point of a circle is perpendicular to the radius through the point of contact].
In right-angled $\triangle OPA$,by Pythagoras theorem:
$OA^2 = OP^2 + PA^2$
$(13)^2 = 5^2 + PA^2$
$169 = 25 + PA^2$
$PA^2 = 144$
$PA = 12\, cm$.
Since tangents from an external point to a circle are equal in length,$AP = AQ = 12\, cm$.
Also,$BP = BR$ and $CR = CQ$ (tangents from points $B$ and $C$ respectively).
Perimeter of $\triangle ABC = AB + BC + CA$
$= AB + (BR + RC) + CA$
$= AB + BP + CQ + CA$
$= (AB + BP) + (CQ + CA)$
$= AP + AQ$
$= 12 + 12 = 24\, cm$.

(C) In the quadrilateral $OXPY$,the radii $OX$ and $OY$ are perpendicular to the tangents $PX$ and $PY$ respectively at the points of contact.
Therefore,$\angle OXP = 90^\circ$ and $\angle OYP = 90^\circ$.
The sum of the angles in a quadrilateral is $360^\circ$.
$\angle XOY + \angle OXP + \angle XPY + \angle OYP = 360^\circ$
$100^\circ + 90^\circ + \angle XPY + 90^\circ = 360^\circ$
$280^\circ + \angle XPY = 360^\circ$
$\angle XPY = 360^\circ - 280^\circ = 80^\circ$.
Since $OP$ is the angle bisector of $\angle XPY$,we have:
$m \angle XPO = \frac{1}{2} m \angle XPY$
$m \angle XPO = \frac{1}{2} (80^\circ) = 40^\circ$.

(D) The radius $OY$ is perpendicular to the tangent $PY$ at the point of contact $Y$.
Therefore,$\Delta PYO$ is a right-angled triangle with $\angle OYP = 90^{\circ}$.
In $\Delta PYO$,by the Pythagorean theorem:
$OP^2 = OY^2 + PY^2$
Given $OP = 30$ and $PY = 24$,and $OY = r$:
$30^2 = r^2 + 24^2$
$900 = r^2 + 576$
$r^2 = 900 - 576$
$r^2 = 324$
$r = \sqrt{324} = 18$
Thus,the radius $r$ is $18$.

(240/17) Let $R$ be the intersection of $OP$ and $AB$. Since $OP$ is the perpendicular bisector of the chord $AB$,$OP \perp AB$ at $R$.
In right-angled $\Delta OAP$,$\angle OAP = 90^{\circ}$.
By Pythagoras theorem,$OP^2 = OA^2 + AP^2$.
$(17)^2 = (8)^2 + AP^2 \implies 289 = 64 + AP^2 \implies AP^2 = 225 \implies AP = 15$.
In $\Delta OAP$,$AR$ is the altitude to the hypotenuse $OP$.
Using the property of right triangles,$OA^2 = OR \cdot OP$.
$(8)^2 = OR \cdot 17 \implies OR = \frac{64}{17}$.
In right-angled $\Delta OAR$,$AR^2 = OA^2 - OR^2$.
$AR^2 = 8^2 - \left(\frac{64}{17}\right)^2 = 64 - \frac{4096}{289} = \frac{18496 - 4096}{289} = \frac{14400}{289}$.
$AR = \sqrt{\frac{14400}{289}} = \frac{120}{17}$.
Since $OP$ bisects $AB$,$AB = 2 \cdot AR = 2 \cdot \left(\frac{120}{17}\right) = \frac{240}{17}$.

(40/3) Let $O$ be the center of the circle with radius $r = 10$. Let $M$ be the midpoint of chord $AB$. Since $AB = 16$,$AM = MB = 8$. In $\triangle OMA$,by the Pythagorean theorem,$OM = \sqrt{OA^2 - AM^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$. Let $PA = x$. In $\triangle OAP$,$\angle OAP = 90^\circ$ (radius is perpendicular to the tangent). Let $\angle AOP = \theta$. Then $\triangle OMA \sim \triangle OAP$ by $AA$ similarity. Thus,$\frac{PA}{AM} = \frac{OA}{OM}$,which gives $\frac{x}{8} = \frac{10}{6}$. Solving for $x$,we get $x = \frac{80}{6} = \frac{40}{3}$.

(A) In the circle $\odot(O, r)$,$PA$ and $PB$ are tangents drawn from an external point $P$ to the points of contact $A$ and $B$.
By the properties of tangents,the radius is perpendicular to the tangent at the point of contact. Therefore,$\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.
In the quadrilateral $OAPB$,the sum of the interior angles is $360^\circ$.
Thus,$\angle AOB + \angle OAP + \angle OBP + \angle APB = 360^\circ$.
Substituting the given values: $100^\circ + 90^\circ + 90^\circ + \angle APB = 360^\circ$.
$280^\circ + \angle APB = 360^\circ$,which gives $\angle APB = 80^\circ$.
The line $OP$ bisects $\angle APB$. Therefore,$m\angle OPB = \frac{1}{2} \angle APB = \frac{1}{2} \times 80^\circ = 40^\circ$.

(D) $1$. In $\triangle OBP$,since $PB$ is a tangent to the circle at $B$,the radius $OB$ is perpendicular to the tangent $PB$. Thus,$\angle OBP = 90^\circ$.
$2$. In the right-angled triangle $\triangle OBP$,the sum of angles is $180^\circ$. Therefore,$\angle POB + \angle OBP + \angle OPB = 180^\circ$.
$3$. Substituting the known values: $\angle POB + 90^\circ + 35^\circ = 180^\circ$,which gives $\angle POB = 180^\circ - 125^\circ = 55^\circ$.
$4$. Since the tangents from an external point are equally inclined to the line joining the center to that point,$\triangle OAP \cong \triangle OBP$. Thus,$\angle AOP = \angle POB = 55^\circ$.
$5$. Therefore,$m\angle AOB = \angle AOP + \angle POB = 55^\circ + 55^\circ = 110^\circ$.

(C) Let the center of the circle be $O$ and the two radii be $OA$ and $OB$. The angle between the radii is $\angle AOB = 48^{\circ}$.
Tangents are drawn at points $A$ and $B$. Let these tangents meet at point $P$.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,we have $\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.
In the quadrilateral $OAPB$,the sum of the interior angles is $360^{\circ}$.
Therefore,$\angle AOB + \angle OAP + \angle OBP + \angle APB = 360^{\circ}$.
Substituting the known values: $48^{\circ} + 90^{\circ} + 90^{\circ} + \angle APB = 360^{\circ}$.
$228^{\circ} + \angle APB = 360^{\circ}$.
$\angle APB = 360^{\circ} - 228^{\circ} = 132^{\circ}$.
Thus,the angle between the tangents is $132^{\circ}$.

(D) From the properties of tangents drawn from an external point to a circle,we know that the lengths of tangents drawn from an external point to a circle are equal.
Therefore,$PA = PB = 8$.
In $\triangle PAB$,since $PA = PB$,it is an isosceles triangle.
Thus,the angles opposite to equal sides are equal,so $m \angle PBA = m \angle PAB = 60^\circ$.
Now,using the angle sum property of a triangle,$m \angle APB + m \angle PAB + m \angle PBA = 180^\circ$.
$m \angle APB + 60^\circ + 60^\circ = 180^\circ$.
$m \angle APB = 180^\circ - 120^\circ = 60^\circ$.
Since all angles of $\triangle PAB$ are $60^\circ$,it is an equilateral triangle.
Therefore,$AB = PA = PB = 8$.

(C) According to the theorem of tangents drawn from an external point to a circle,the lengths of tangents drawn from an external point to a circle are equal.
Since $P$ is an external point and $\overline{ PA }$ and $\overline{ PB }$ are tangents to the circle $\odot( O , r)$,it follows that $PA = PB$.

(B) According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.
Therefore,$PA = PB$.
Given that $PA = 8$,it follows that $PB = 8$.

(D) In the quadrilateral $OAPB$,the radius is perpendicular to the tangent at the point of contact.
Therefore,$\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.
The sum of the angles in a quadrilateral is $360^{\circ}$.
Thus,$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^{\circ}$.
Substituting the known values: $\angle AOB + 90^{\circ} + 65^{\circ} + 90^{\circ} = 360^{\circ}$.
$\angle AOB + 245^{\circ} = 360^{\circ}$.
$\angle AOB = 360^{\circ} - 245^{\circ} = 115^{\circ}$.

(D) The line segment $OP$ bisects the angle $\angle APB$.
Therefore,$m \angle OPB = \frac{1}{2} m \angle APB = \frac{1}{2} (70^\circ) = 35^\circ$.
Since $PB$ is a tangent to the circle at point $B$,the radius $OB$ is perpendicular to the tangent $PB$.
Thus,$m \angle OBP = 90^\circ$.
In the right-angled triangle $\Delta OBP$,the sum of the angles is $180^\circ$.
$m \angle POB + m \angle OBP + m \angle OPB = 180^\circ$.
$m \angle POB + 90^\circ + 35^\circ = 180^\circ$.
$m \angle POB + 125^\circ = 180^\circ$.
$m \angle POB = 180^\circ - 125^\circ = 55^\circ$.

(B) tangent to a circle is a line that intersects the circle at exactly one point.
From any point $P$ lying in the exterior of a circle,exactly two tangents can be drawn to the circle.
These two tangents touch the circle at two distinct points.
Therefore,the maximum number of tangents that can be drawn from an external point to a circle is $2$.

(C) Let the tangent $\overleftrightarrow{PQ}$ touch the circle at point $R$.
Since the lengths of tangents drawn from an external point to a circle are equal,we have:
$PB = PR$ and $QC = QR$.
The perimeter of $\Delta APQ = AP + PQ + AQ$.
$= AP + (PR + RQ) + AQ$.
$= AP + PB + QC + AQ$ (since $PR = PB$ and $RQ = QC$).
$= (AP + PB) + (AQ + QC)$.
$= AB + AC$.
Since $AB = AC$ (tangents from an external point $A$),we have $AB = AC = 6$.
Therefore,the perimeter of $\Delta APQ = 6 + 6 = 12$.

(C) Given that $\overline{ PA }$ and $\overline{ PB }$ are tangents to the circle $\odot( O , r)$ from an external point $P$.
According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.
Therefore,$\overline{ PA } = \overline{ PB }$.
In $\Delta PAB$,since $\overline{ PA } = \overline{ PB }$,the triangle is an isosceles triangle.
In an isosceles triangle,the angles opposite to the equal sides are equal.
Thus,$m \angle PAB = m \angle PBA$.
Given that $m \angle PAB = 60^{\circ}$,it follows that $m \angle PBA = 60^{\circ}$.

(B) When two circles touch each other externally,they have one direct common tangent passing through the point of contact and two transverse common tangents.
Therefore,a total of $3$ common tangents can be drawn to the two circles.