Construction of Tangent to a Circle

In Euclidean geometry, every compass-and-straightedge construction must be based on an unshakeable mathematical theorem. For circle tangents, the governing theorem is Theorem 40 and its converse:

Therefore, to construct a tangent at any point, the draftsman's fundamental task is to construct a perpendicular line ($90^\circ$ angle) to the radius at the point of contact.

The WBBSE secondary syllabus specifies two classical tangent construction problems:

• Construction 1: Constructing a tangent to a circle at a given point lying on the circumference.
• Construction 2: Constructing two tangents to a circle from an external point lying outside the circle.

Problem: Draw a circle of radius $r$ with center $O$. Take any point $P$ on its circumference. Construct a tangent to the circle at point $P$.

1. Step 1 (Draw Circle): Fix point $O$ on the paper as center. With a compass set to radius $r$, draw the given circle.
2. Step 2 (Mark Point & Draw Radius): Mark any point $P$ on the circumference. Using a straightedge ruler, draw the line segment $OP$ and extend it slightly outward beyond $P$ to form ray $OX$.
3. Step 3 (Construct 90° Angle at P):
   • Place the compass needle at point $P$. With any convenient radius, draw an arc cutting the ray $OX$ at point $E$ and the segment $OP$ at point $F$.
   • With the compass needle at $E$ and radius equal to the previous arc, cut an arc at point $G$ ($60^\circ$ mark).
   • With the needle at $G$ and same radius, cut another arc along the curve at point $H$ ($120^\circ$ mark).
   • With the needle at $G$ and then at $H$ (keeping the same radius), draw two intersecting arcs above $P$ that cross each other at point $K$.
4. Step 4 (Draw Tangent Line): Using a straightedge, draw a line passing through point $P$ and intersection point $K$. Extend this line in both directions to form line $AB$.
5. Conclusion: The straight line $AB$ is the required tangent to the circle at point $P$.

Problem: Draw a circle of radius $r$ with center $O$. Take an external point $P$ at a distance $d$ from the center ($d > r$). Construct two tangents from point $P$ to the circle.

1. Step 1 (Draw Circle & Locate External Point): Draw a circle with center $O$ and radius $r$. With a straightedge ruler, draw a straight line from $O$ and mark point $P$ such that $OP = d$.
2. Step 2 (Bisect Line Segment OP):
   • Place the compass needle at point $O$. Open the compass to a radius strictly greater than half of $OP$ ($> rac{1}{2}OP$).
   • Draw two arcs, one above the segment $OP$ and one below $OP$.
   • Keeping the compass radius unchanged, place the needle at point $P$ and draw two arcs intersecting the previous arcs at points $X$ and $Y$.
   • Join $X$ and $Y$ with a thin dashed line. The line $XY$ intersects $OP$ at point $M$. Point $M$ is the exact midpoint of OP ($OM = MP$).
3. Step 3 (Draw Auxiliary Semicircle / Circle):
   • Place the compass needle at the midpoint $M$. Set the compass radius equal to $MO$ (which equals $MP$).
   • Draw a semicircle (or full circle) passing through both center $O$ and external point $P$.
4. Step 4 (Mark Contact Points A and B): The auxiliary semicircle/circle will intersect the original circle at two points. Mark the upper intersection point as $A$ and the lower intersection point as $B$.
5. Step 5 (Draw Tangents): Using a straightedge, join point $P$ to point $A$ and extend it. Join point $P$ to point $B$ and extend it.
6. Conclusion: The lines $PA$ and $PB$ are the two required tangents to the circle from external point $P$.

Proof: Join the radii $OA$ and $OB$ with thin dashed lines.

1. In the auxiliary circle/semicircle, $OP$ is the diameter because $M$ is the center and $MO = MP$ is the radius.
2. Point $A$ lies on the circumference of this auxiliary circle.
3. Therefore, $ngle OAP$ is an angle inscribed in a semicircle.
4. By Thales' Theorem (WBBSE Class 10 Chapter 7, Theorem 37), the angle in a semicircle is a right angle: $$\mathbf{ngle OAP = 90^\circ \implies OA \perp PA}$$
5. Now, in the original circle, $OA$ is a radius and point $A$ lies on the circumference.
6. Since $PA$ passes through the endpoint $A$ of radius $OA$ and is perpendicular to $OA$, by the converse of Theorem 40, $PA$ must be a tangent to the circle at point $A$.
7. By exactly identical reasoning, $ngle OBP = 90^\circ \implies OB \perp PB$, proving that $PB$ is also a tangent at point $B$. (Hence Proved)

In right-angled triangle $ riangle OAP$, hypotenuse is $OP$ and radius is $OA = r$. By Pythagoras' Theorem:

Students should always calculate this theoretical value using Pythagoras' Theorem and compare it to their measured ruler length. For example, if $r = 3 ext{ cm}$ and $OP = 5 ext{ cm}$, the measured tangent length must be exactly $\sqrt{5^2 - 3^2} = \sqrt{16} = 4.0 ext{ cm}$. Examiners allow a tolerance of $\pm 1 ext{ mm}$ ($3.9 ext{ cm}$ to $4.1 ext{ cm}$).

Problem: Draw a circle of radius $r$. Construct two tangents to this circle such that the angle between them is $ heta$ (for example, $ heta = 60^\circ$ or $45^\circ$).

Key Strategy (The Supplementary Central Angle Rule): We cannot directly construct an angle between tangents at an unknown external point $P$. Instead, we use the property that quadrilateral $OAPB$ is cyclic ($ngle OAP = 90^\circ, ngle OBP = 90^\circ$): $$ngle APB + ngle AOB = 180^\circ \implies \mathbf{ngle AOB = 180^\circ - heta}$$

1. Draw circle with center $O$ and radius $r$. Draw any radius $OA$.
2. At center $O$, construct angle $ngle AOB = 180^\circ - heta$ using a compass. For example, if $ heta = 60^\circ$, construct $ngle AOB = 180^\circ - 60^\circ = 120^\circ$. Draw radius $OB$.
3. At endpoint $A$, erect a perpendicular line $AP$ to $OA$ ($ngle OAP = 90^\circ$).
4. At endpoint $B$, erect a perpendicular line $BP$ to $OB$ ($ngle OBP = 90^\circ$).
5. The two perpendiculars will intersect at point $P$. By quadrilateral angle sum, $ngle APB = 360^\circ - (90^\circ + 90^\circ + 120^\circ) = 60^\circ$.