Construction of Tangents to a Circle Questions in English

(N/A) pair of tangents to the given circle can be constructed as follows:
$1.$ Taking any point $O$ of the given plane as centre,draw a circle of $6 \, cm$ radius. Locate a point $P$,$10 \, cm$ away from $O$.
Join $OP$.
$2.$ Bisect $OP$. Let $M$ be the mid-point of $PO$.
$3.$ Taking $M$ as centre and $MO$ as radius,draw a circle.
$4.$ Let this circle intersect the previous circle at points $Q$ and $R$.
$5.$ Join $PQ$ and $PR$. $PQ$ and $PR$ are the required tangents.
The lengths of tangents $PQ$ and $PR$ are $8 \, cm$ each.
Justification:
The construction can be justified by proving that $PQ$ and $PR$ are the tangents to the circle (whose centre is $O$ and radius is $6 \, cm$). For this,join $OQ$ and $OR$.
$\angle PQO$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
$\therefore \angle PQO = 90^{\circ}$
$\Rightarrow OQ \perp PQ$
Since $OQ$ is the radius of the circle,$PQ$ has to be a tangent to the circle. Similarly,$PR$ is a tangent to the circle.

(N/A) Tangents on the given circle can be drawn as follows:
$1.$ Draw a circle of $4 \, cm$ radius with centre as $O$ on the given plane.
$2.$ Draw a circle of $6 \, cm$ radius taking $O$ as its centre. Locate a point $P$ on this circle and join $OP.$
$3.$ Bisect $OP.$ Let $M$ be the mid-point of $PO.$
$4.$ Taking $M$ as its centre and $MO$ as its radius,draw a circle. Let it intersect the given circle at the points $Q$ and $R$.
$5.$ Join $PQ$ and $PR.$ $PQ$ and $PR$ are the required tangents.
It can be observed that $PQ$ and $PR$ are of length $4.47 \, cm$ each.
In $\triangle PQO$:
Since $PQ$ is a tangent,$\angle PQO = 90^{\circ}$.
$PO = 6 \, cm$ (hypotenuse),
$QO = 4 \, cm$ (radius).
Applying the Pythagoras theorem in $\triangle PQO$,we obtain:
$PQ^2 + QO^2 = PO^2$
$PQ^2 + (4)^2 = (6)^2$
$PQ^2 + 16 = 36$
$PQ^2 = 36 - 16$
$PQ^2 = 20$
$PQ = \sqrt{20} = 2\sqrt{5} \approx 4.47 \, cm$.
Justification:
The construction can be justified by proving that $PQ$ and $PR$ are the tangents to the circle (whose centre is $O$ and radius is $4 \, cm$). For this,join $OQ$ and $OR$.
Since $\angle OQP$ is an angle in a semi-circle,$\angle OQP = 90^{\circ}$.
$\Rightarrow OQ \perp PQ$.
Since $OQ$ is the radius of the circle,$PQ$ must be a tangent to the circle. Similarly,$PR$ is a tangent to the circle.

(N/A) The tangents can be constructed on the given circle as follows:
$1.$ Taking any point $O$ on the given plane as the centre,draw a circle of $3 \, cm$ radius.
$2.$ Take one of its diameters and extend it on both sides. Locate two points $P$ and $Q$ on this diameter such that $OP = OQ = 7 \, cm$.
$3.$ Bisect $OP$ and $OQ$. Let $T$ and $U$ be the mid-points of $OP$ and $OQ$ respectively.
$4.$ Taking $T$ and $U$ as centres and $TO$ and $UO$ as radii,draw two circles. These two circles will intersect the original circle at points $V, W$ and $X, Y$ respectively. Join $PV, PW, QX,$ and $QY$. These are the required tangents.
Justification:
The construction can be justified by proving that $PV, PW, QX,$ and $QY$ are the tangents to the circle (whose centre is $O$ and radius is $3 \, cm$). For this,join $OV, OW, OX,$ and $OY$.
$\angle PVO$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
$\therefore \angle PVO = 90^{\circ}$
$\Rightarrow OV \perp PV$
Since $OV$ is the radius of the circle,$PV$ must be a tangent to the circle.
Similarly,it can be shown that $PW, QX,$ and $QY$ are the tangents to the circle.

(N/A) The tangents can be constructed in the following manner:
$1.$ Draw a circle of radius $5\, cm$ with center $O$.
$2.$ Take a point $A$ on the circumference of the circle and join $OA$. Draw a perpendicular to $OA$ at point $A$.
$3.$ Draw a radius $OB$ making an angle of $120^{\circ} (180^{\circ} - 60^{\circ})$ with $OA$.
$4.$ Draw a perpendicular to $OB$ at point $B$. Let both the perpendiculars intersect at point $P$. $PA$ and $PB$ are the required tangents inclined at an angle of $60^{\circ}$.
Justification:
The construction can be justified by proving that $\angle APB = 60^{\circ}$.
By our construction:
$\angle OAP = 90^{\circ}$ (Tangent is perpendicular to the radius at the point of contact)
$\angle OBP = 90^{\circ}$ (Tangent is perpendicular to the radius at the point of contact)
And $\angle AOB = 120^{\circ}$
We know that the sum of all interior angles of a quadrilateral is $360^{\circ}$.
In quadrilateral $OAPB$:
$\angle OAP + \angle AOB + \angle OBP + \angle APB = 360^{\circ}$
$90^{\circ} + 120^{\circ} + 90^{\circ} + \angle APB = 360^{\circ}$
$300^{\circ} + \angle APB = 360^{\circ}$
$\angle APB = 360^{\circ} - 300^{\circ} = 60^{\circ}$
This justifies the construction.

(N/A) The tangents can be constructed on the given circles as follows:
$1.$ Draw a line segment $AB$ of $8 \, cm$. Taking $A$ and $B$ as centres,draw two circles of $4 \, cm$ and $3 \, cm$ radius respectively.
$2.$ Bisect the line $AB$. Let the mid-point of $AB$ be $C$. Taking $C$ as centre,draw a circle of radius $AC$ (or $BC$),which will intersect the circles at points $P, Q, R,$ and $S$. Join $BP, BQ, AS,$ and $AR$. These are the required tangents.
Justification:
The construction can be justified by proving that $AS$ and $AR$ are the tangents to the circle (whose centre is $B$ and radius is $3 \, cm$) and $BP$ and $BQ$ are the tangents to the circle (whose centre is $A$ and radius is $4 \, cm$). For this,join $AP, AQ, BS,$ and $BR$.
$\angle ASB$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
$\therefore \angle ASB = 90^{\circ}$
$\Rightarrow BS \perp AS$
Since $BS$ is the radius of the circle,$AS$ must be a tangent to the circle. Similarly,$AR, BP,$ and $BQ$ are the tangents.

(N/A) Construction steps:
$1$. Construct $\triangle ABC$ with $AB = 6 \, cm$,$BC = 8 \, cm$ and $\angle B = 90^{\circ}$.
$2$. Draw $BD \perp AC$. Since $\angle BDC = 90^{\circ}$,the circle passing through $B, C, D$ will have $BC$ as its diameter because $\angle BDC = 90^{\circ}$ subtends a right angle at $D$.
$3$. Let $O$ be the midpoint of $BC$. Draw a circle with center $O$ and radius $OB = OC = 4 \, cm$. This circle passes through $B, C$ and $D$ (since $\angle BDC = 90^{\circ}$).
$4$. To construct tangents from $A$ to this circle,join $AO$. Bisect $AO$ at $M$. Taking $M$ as center and $MA$ as radius,draw a circle. Let this circle intersect the circle with center $O$ at $B$ and $E$. Join $AE$. Thus,$AB$ and $AE$ are the required tangents.
Justification:
Join $OE$. Since $AB$ is a tangent to the circle at $B$,$\angle ABO = 90^{\circ}$. Since $AE$ is a tangent to the circle at $E$,$\angle AEO = 90^{\circ}$. In the circle with center $O$,$OB$ and $OE$ are radii. Since $AB$ and $AE$ are tangents from an external point $A$,$AB = AE$.

(N/A) To construct tangents to a circle drawn with a bangle (where the centre is unknown):
$1$. Draw two non-parallel chords $BC$ and $CD$.
$2$. Draw the perpendicular bisectors of $BC$ and $CD$. The point where these bisectors intersect is the centre $E$ of the circle.
$3$. Take a point $A$ outside the circle. Join $AE$.
$4$. Bisect $AE$ at point $F$. With $F$ as the centre and $FA$ as the radius,draw a circle.
$5$. Let this circle intersect the original circle at points $B$ and $G$. Join $AB$ and $AG$.
$AB$ and $AG$ are the required tangents.
Justification:
Join $EB$ and $EG$. $\angle ABE$ is an angle in a semi-circle,so $\angle ABE = 90^{\circ}$. Since $EB$ is a radius,$AB$ must be a tangent. Similarly,$\angle AGE = 90^{\circ}$,so $AG$ is a tangent.

(B) Let the angle between the two tangents be $\theta = 35^{\circ}.$
In the quadrilateral formed by the center of the circle,the two points of contact,and the intersection point of the tangents,the angles at the points of contact are $90^{\circ}$ each (since the radius is perpendicular to the tangent).
Thus,the sum of the angles in the quadrilateral is $360^{\circ}.$
Let the angle between the radii be $\alpha.$
Then,$\alpha + 90^{\circ} + \theta + 90^{\circ} = 360^{\circ}.$
$\alpha + \theta = 180^{\circ}.$
$\alpha = 180^{\circ} - 35^{\circ} = 145^{\circ}.$

(D) Let the center of the circle be $O$ and the intersection point of the two tangents be $P$. Let the points of contact of the tangents be $A$ and $B$.
In the quadrilateral $OAPB$,the angles at the points of contact $A$ and $B$ are $90^{\circ}$ each (since the tangent is perpendicular to the radius at the point of contact).
The sum of the angles in a quadrilateral is $360^{\circ}$.
Therefore,$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^{\circ}$.
Substituting the known values: $\angle AOB + 90^{\circ} + 60^{\circ} + 90^{\circ} = 360^{\circ}$.
$\angle AOB + 240^{\circ} = 360^{\circ}$.
$\angle AOB = 360^{\circ} - 240^{\circ} = 120^{\circ}$.
Thus,the angle between the two radii should be $120^{\circ}$.

(A) True.
In a circle,the angle between the two tangents drawn from an external point and the angle subtended by the line segments joining the points of contact to the center are supplementary (i.e.,they add up to $180^{\circ}$).
Let the angle between the tangents be $\theta$ and the angle at the center be $\phi$.
We know that $\theta + \phi = 180^{\circ}$.
Since the angle between the tangents is given as $170^{\circ}$,the angle at the center would be $180^{\circ} - 170^{\circ} = 10^{\circ}$.
Since $10^{\circ} > 0^{\circ}$,it is geometrically possible to construct such a pair of tangents.

(N/A) Given,a point $M^{\prime}$ is at a distance of $6\, cm$ from the centre of a circle of radius $4\, cm$.
Steps of construction:
$1.$ Draw a circle of radius $4\, cm$. Let the centre of this circle be $O$.
$2.$ Join $OM^{\prime}$ and bisect it. Let $M$ be the mid-point of $OM^{\prime}$.
$3.$ Taking $M$ as the centre and $MO$ as the radius,draw a circle to intersect the circle $(O, 4\, cm)$ at two points,$P$ and $Q$.
$4.$ Join $PM^{\prime}$ and $QM^{\prime}$. $PM^{\prime}$ and $QM^{\prime}$ are the required tangents from $M^{\prime}$ to the circle $C(O, 4\, cm)$.

(N/A) Given: Two concentric circles with centre $O$ and radii $3\, cm$ and $5\, cm$. We need to draw a pair of tangents from a point $P$ on the outer circle to the inner circle.
Steps of construction:
$1.$ Draw two concentric circles with centre $O$ and radii $3\, cm$ and $5\, cm$.
$2.$ Take any point $P$ on the outer circle. Join $OP$.
$3.$ Bisect $OP$. Let $M'$ be the midpoint of $OP$.
$4.$ Taking $M'$ as the centre and $OM'$ as the radius,draw a dotted circle which intersects the inner circle at points $M$ and $P'$.
$5.$ Join $PM$ and $PP'$. Thus,$PM$ and $PP'$ are the required tangents.
$6.$ On measuring $PM$ and $PP'$,we find that $PM = PP' = 4\, cm$.
Actual calculation:
In right-angled $\triangle OMP$,$\angle PMO = 90^{\circ}$.
By Pythagoras theorem: $OP^2 = PM^2 + OM^2$
$PM^2 = OP^2 - OM^2$
$PM^2 = (5)^2 - (3)^2 = 25 - 9 = 16$
$PM = \sqrt{16} = 4\, cm$.
Hence,the length of both tangents is $4\, cm$.

(N/A) Steps of construction:
$1.$ Draw a circle with centre $O$ and radius $4 \, cm$.
$2.$ Draw a radius $OA$ and extend it to $B$ such that $OA = AB = 4 \, cm$. Thus,$OB = 8 \, cm$.
$3.$ Draw a circle with centre $A$ and radius $AO = 4 \, cm$. Let this circle intersect the original circle at points $P$ and $Q$.
$4.$ Join $BP$ and $BQ$. $BP$ and $BQ$ are the required tangents.
Justification:
In $\triangle OAP$,$OA = OP = 4 \, cm$ (radii) and $AP = 4 \, cm$ (radius of circle with centre $A$).
Since all sides are equal,$\triangle OAP$ is an equilateral triangle.
Therefore,$\angle OAP = 60^{\circ}$.
Since $OAB$ is a straight line,$\angle BAP = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
In $\triangle BAP$,$BA = AP = 4 \, cm$. Thus,$\triangle BAP$ is an isosceles triangle.
Therefore,$\angle ABP = \angle APB = (180^{\circ} - 120^{\circ}) / 2 = 30^{\circ}$.
Similarly,$\angle ABQ = 30^{\circ}$.
Thus,$\angle PBQ = \angle ABP + \angle ABQ = 30^{\circ} + 30^{\circ} = 60^{\circ}$.
Measurement:
The distance between the centre $O$ and the point of intersection $B$ is $OB = OA + AB = 4 \, cm + 4 \, cm = 8 \, cm$.

(N/A) Data: Draw a circle with centre $A$ and radius $5 \, cm$. Take a point $B$ in its exterior such that $AB = 8 \, cm$.
To construct: Through point $B$,draw two tangents to the circle with centre $A$ and radius $5 \, cm$.
Steps of construction:
$1$. Draw a circle with centre $A$ and radius $5 \, cm$ and mark a point $B$ outside the circle such that $AB = 8 \, cm$.
$2$. Draw the line segment $\overline{AB}$.
$3$. Find the midpoint $M$ of $\overline{AB}$ by constructing its perpendicular bisector.
$4$. With $M$ as the centre and $MA$ as the radius,draw a circle that intersects the original circle at points $X$ and $Y$.
$5$. Draw the rays $\overrightarrow{BX}$ and $\overrightarrow{BY}$.
Thus,$\overleftrightarrow{BX}$ and $\overleftrightarrow{BY}$ are the required tangents.
Calculation: In $\triangle AXB$,$\angle AXB = 90^\circ$ (angle in a semicircle). By Pythagoras theorem,$AB^2 = AX^2 + BX^2$.
$8^2 = 5^2 + BX^2 \implies 64 = 25 + BX^2 \implies BX^2 = 39 \implies BX = \sqrt{39} \approx 6.24 \, cm$.
Therefore,the length of the tangents is approximately $6.24 \, cm$.

(N/A) Data: Draw a circle with the help of a circular bangle and take point $A$ in the exterior of the circle.
To construct: Through $A$,tangents are to be drawn to the circle.
$(1)$ Draw a circle with the help of a circular bangle and take point $A$ in the exterior of the circle.
$(2)$ Draw two non-parallel chords $\overline{PQ}$ and $\overline{RS}$ in this circle.
$(3)$ Draw the perpendicular bisectors of $\overline{PQ}$ and $\overline{RS}$ to intersect each other at $O$. Here,$O$ is the centre of the circle drawn with the help of the circular bangle.
$(4)$ Draw $\overline{OA}$.
$(5)$ Obtain the midpoint $M$ of $\overline{OA}$ by constructing the perpendicular bisector of $\overline{OA}$.
$(6)$ Draw a circle with centre $M$ and radius $MA$ to intersect the first circle at $X$ and $Y$.
$(7)$ Draw rays $\overrightarrow{AX}$ and $\overrightarrow{AY}$.
Thus,$\overleftrightarrow{AX}$ and $\overleftrightarrow{AY}$ are the required tangents.

(N/A) Data: Draw $\overline{AB}$ such that $AB = 10 \, cm$. Draw $\odot(A, 3 \, cm)$ and $\odot(B, 4 \, cm)$.
To construct: From the centre of each circle,tangents are to be drawn to the other circle.
Steps of construction:
$(1)$ Draw $\overline{AB}$ such that $AB = 10 \, cm$. Draw $\odot(A, 3 \, cm)$ and $\odot(B, 4 \, cm)$.
$(2)$ Obtain the midpoint $M$ of $\overline{AB}$ by constructing its perpendicular bisector.
$(3)$ Draw $\odot(M, MA)$ to intersect $\odot(A, 3 \, cm)$ at $S$ and $T$,and $\odot(B, 4 \, cm)$ at $X$ and $Y$.
$(4)$ Draw $\overrightarrow{AX}$ and $\overrightarrow{AY}$ as well as $\overrightarrow{BS}$ and $\overrightarrow{BT}$.
Thus,$\overleftrightarrow{AX}$ and $\overleftrightarrow{AY}$ are the tangents from $A$ to $\odot(B, 4 \, cm)$ and $\overleftrightarrow{BS}$ and $\overleftrightarrow{BT}$ are the tangents from $B$ to $\odot(A, 3 \, cm)$.

(N/A) Data: Draw $\odot(P, 4 \text{ cm})$.
To construct: Draw a pair of tangents to $\odot(P, 4 \text{ cm})$ such that the measure of the angle between the tangents at their point of intersection $A$ is $60^\circ$.
$(1)$ Draw $\odot(P, 4 \text{ cm})$ and two radii $\overline{PR}$ and $\overline{PQ}$ such that $m\angle RPQ = 180^\circ - 60^\circ = 120^\circ$.
$(2)$ Through $R$,draw a line perpendicular to $\overline{PR}$.
$(3)$ Through $Q$,draw a line perpendicular to $\overline{PQ}$.
$(4)$ Let the lines drawn in step $(2)$ and step $(3)$ intersect at $A$.
Thus,$\overleftrightarrow{AR}$ and $\overleftrightarrow{AQ}$ are the required tangents such that the measure of the angle between them is $60^\circ$.

(N/A) Steps of construction:
$1$. Draw a circle with center $P$ and radius $3\, cm$.
$2$. Mark a point $R$ outside the circle such that the distance $PR = 6\, cm$.
$3$. Find the midpoint $M$ of the line segment $PR$ by drawing its perpendicular bisector.
$4$. With $M$ as the center and $MP$ (or $MR$) as the radius,draw a second circle.
$5$. Let this circle intersect the original circle at points $A$ and $B$.
$6$. Join $RA$ and $RB$. These are the required tangents to the circle $\odot(P, 3\, cm)$ from point $R$.

(N/A) Steps of construction:
$1$. Draw a circle with center $O$ and radius $4 \text{ cm}$.
$2$. Mark a point $X$ outside the circle such that the distance $OX = 7 \text{ cm}$.
$3$. Draw the perpendicular bisector of the line segment $OX$. Let $M$ be the midpoint of $OX$.
$4$. With $M$ as the center and $MO$ as the radius,draw a circle. This circle will intersect the original circle at two points,say $P$ and $Q$.
$5$. Join $XP$ and $XQ$. These are the required tangents to the circle from point $X$.

(N/A) Steps of construction:
$1$. Draw a circle with center $P$ and radius $5 \,cm$.
$2$. Mark a point $A$ outside the circle such that the distance $PA = 9 \,cm$.
$3$. Draw the perpendicular bisector of the line segment $PA$. Let $M$ be the midpoint of $PA$.
$4$. With $M$ as the center and $MA$ (or $MP$) as the radius,draw a circle. This circle will intersect the original circle at two points,say $B$ and $C$.
$5$. Join $AB$ and $AC$. Thus,$AB$ and $AC$ are the required pair of tangents to the circle from point $A$.

(N/A) Step $1$: Place a round bowl upside-down on a paper and trace its boundary to draw a circle.
Step $2$: Since the center of the circle is unknown,draw two non-parallel chords $AB$ and $CD$ of the circle.
Step $3$: Draw the perpendicular bisectors of chords $AB$ and $CD$. The point where these bisectors intersect is the center $O$ of the circle.
Step $4$: Mark a point $P$ outside the circle.
Step $5$: Join $OP$ and draw the perpendicular bisector of $OP$. Let $M$ be the midpoint of $OP$.
Step $6$: With $M$ as the center and $MO$ as the radius,draw a circle. Let this circle intersect the original circle at points $Q$ and $R$.
Step $7$: Join $PQ$ and $PR$. $PQ$ and $PR$ are the required tangents to the circle from point $P$.

(N/A) Steps of construction:
$1$. Draw a circle with center $O$ and radius $4\, cm$.
$2$. Draw a diameter $\overline{PQ}$ passing through the center $O$.
$3$. Extend the diameter $\overline{PQ}$ on both sides to locate points $A$ and $B$ such that $A-P-Q$ and $P-Q-B$. Let $OA = OB = 6\, cm$ (or any distance greater than the radius).
$4$. Find the midpoints of $\overline{OA}$ and $\overline{OB}$ by constructing perpendicular bisectors. Let these midpoints be $M_1$ and $M_2$ respectively.
$5$. With $M_1$ as center and $M_1O$ as radius,draw a circle. It intersects the original circle at two points. Join $A$ to these points to get the tangents from $A$.
$6$. Similarly,with $M_2$ as center and $M_2O$ as radius,draw a circle. It intersects the original circle at two points. Join $B$ to these points to get the tangents from $B$.

(N/A) Steps of construction:
$1$. Draw a circle with center $P$ and radius $3 \, cm$.
$2$. Draw a diameter $\overline{ AB }$ passing through $P$.
$3$. Extend the line segment $\overline{ AB }$ on both sides. Mark point $X$ on the side of $A$ such that $PX = 7 \, cm$ and point $Y$ on the side of $B$ such that $PY = 7 \, cm$.
$4$. To draw tangents from $X$,find the midpoint $M_1$ of $\overline{ PX }$. With $M_1$ as center and $M_1P$ as radius,draw a circle. Let it intersect the original circle at points $Q$ and $R$.
$5$. Join $XQ$ and $XR$. These are the required tangents from $X$.
$6$. Similarly,to draw tangents from $Y$,find the midpoint $M_2$ of $\overline{ PY }$. With $M_2$ as center and $M_2P$ as radius,draw a circle. Let it intersect the original circle at points $S$ and $T$.
$7$. Join $YS$ and $YT$. These are the required tangents from $Y$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{ AB } = 9\, cm$.
$2$. With $A$ as the center,draw a circle of radius $3\, cm$. With $B$ as the center,draw a circle of radius $3.5\, cm$.
$3$. To draw tangents from $B$ to $\odot( A, 3\, cm )$:
a. Find the midpoint $M$ of $\overline{ AB }$ by drawing the perpendicular bisector of $\overline{ AB }$.
b. With $M$ as the center and $MA$ as the radius,draw a circle. This circle intersects $\odot( A, 3\, cm )$ at points $P$ and $Q$.
c. Join $BP$ and $BQ$. These are the required tangents from $B$ to the circle centered at $A$.
$4$. To draw tangents from $A$ to $\odot( B, 3.5\, cm )$:
a. With $M$ as the center and $MB$ as the radius,draw a circle (which is the same as the one drawn in step $3b$). This circle intersects $\odot( B, 3.5\, cm )$ at points $R$ and $S$.
b. Join $AR$ and $AS$. These are the required tangents from $A$ to the circle centered at $B$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{ XY } = 11\, cm$.
$2$. Draw a circle with center $X$ and radius $4\, cm$,and another circle with center $Y$ and radius $3\, cm$.
$3$. Find the midpoint $M$ of $\overline{ XY }$ by drawing the perpendicular bisector of $\overline{ XY }$.
$4$. With $M$ as the center and $MX$ (or $MY$) as the radius,draw a circle. This circle will intersect the circle with center $X$ at points $A$ and $B$,and the circle with center $Y$ at points $C$ and $D$.
$5$. Join $XA$ and $XB$ to get the tangents from $X$ to the circle centered at $Y$. (Note: Since the circles are separate,tangents from the center of one circle to the other are drawn by finding the intersection points of the auxiliary circle with the target circle).
$6$. Join $YC$ and $YD$ to get the tangents from $Y$ to the circle centered at $X$.

(N/A) $1$. Draw a circle with center $P$ and radius $5 \, cm$.
$2$. Draw a radius $PA$ and another radius $PB$ such that $\angle APB = 120^\circ$ (since the angle between tangents is $60^\circ$,the angle at the center is $180^\circ - 60^\circ = 120^\circ$).
$3$. At point $A$,construct a line perpendicular to $PA$.
$4$. At point $B$,construct a line perpendicular to $PB$.
$5$. Let these two perpendicular lines intersect at point $Q$. Then $QA$ and $QB$ are the required tangents,and $\angle AQB = 60^\circ$.

(N/A) $1$. Draw a circle with center $P$ and radius $4\, cm$.
$2$. Draw two perpendicular diameters $AB$ and $CD$ of the circle,intersecting at the center $P$.
$3$. At points $A, B, C,$ and $D$,draw tangents to the circle using a set square or by constructing $90^{\circ}$ angles at each point.
$4$. Let the tangents at $A$ and $C$ intersect at point $Q$. Since the radii $PA$ and $PC$ are perpendicular,the quadrilateral $PAQC$ is a square (as all angles are $90^{\circ}$ and adjacent sides are equal to the radius).
$5$. Thus,the tangents at $A$ and $C$ are perpendicular to each other at point $Q$.

(N/A) $1$. Draw a circle with center $P$ and radius $5 \, cm$.
$2$. Draw a radius $PA$ and another radius $PB$ such that $\angle APB = 180^\circ - 120^\circ = 60^\circ$. This is because the angle between the tangents and the angle between the radii at the points of contact are supplementary $(180^\circ)$.
$3$. At point $A$,construct a line perpendicular to $PA$.
$4$. At point $B$,construct a line perpendicular to $PB$.
$5$. Let these two perpendicular lines intersect at point $Q$. The lines $QA$ and $QB$ are the required tangents,and $\angle AQB = 120^\circ$.