Theorems Related to Tangent to a Circle

Consider a circle with center $O$ and radius $r$ in a plane, and a straight line $AB$ in the same plane. There are exactly three mutually exclusive relative positions:

• Non-intersecting line: The line $AB$ has no common point with the circle. The perpendicular distance from center $O$ to line $AB$ is strictly greater than the radius $r$ ($d > r$).
• Secant Line (ছেদক): The line intersects the circle at two distinct points $P$ and $Q$. The chord $PQ$ is the segment intercepted by the circle. Here, the perpendicular distance from center $O$ to line $AB$ is strictly less than the radius ($d < r$).
• Tangent Line (স্পর্শক): The line touches the circle at exactly one unique point $P$. The perpendicular distance from the center $O$ to the line equals the radius ($d = r$).

The single common point $P$ shared by the circle and the tangent line is formally defined as the Point of Contact (স্পর্শবিন্দু). Mathematically, a tangent is the limiting position of a secant when the two points of intersection $P$ and $Q$ move along the circumference and coalesce into a single coincident point ($Q o P$).

Theorem 40: The tangent at any point of a circle is perpendicular to the radius through the point of contact (বৃত্তের কোনো বিন্দুতে স্পর্শক এবং ওই স্পর্শবিন্দুগামী ব্যাসার্ধ পরস্পর লম্বভাবে অবস্থিত)।

Given: Let a circle with center $O$ have a tangent straight line $XY$ touching the circle at point $P$. $OP$ is the radius drawn through the point of contact $P$.

To Prove: $OP \perp XY$ (that is, $OP$ is perpendicular to $XY$).

Construction: Take any point $Q$ on the tangent line $XY$, other than the point of contact $P$. Join $O$ and $Q$. Let the line segment $OQ$ intersect the circle at point $R$.

Proof:

1. Since $XY$ is the tangent touching the circle only at point $P$, every point on $XY$ other than $P$ must lie outside the circle.
2. Therefore, point $Q$ lies strictly outside the circle.
3. Since point $R$ lies on the circumference of the circle on the segment $OQ$, we have: $$OQ = OR + RQ \implies OQ > OR$$
4. Now, both $OP$ and $OR$ are radii of the same circle. Hence, $OP = OR$.
5. Substituting $OP$ for $OR$ in the inequality yields: $$OQ > OP$$
6. This inequality holds true for every point $Q$ on the tangent line $XY$ except point $P$.
7. Hence, the segment $OP$ is the shortest distance from the center $O$ to any point on the line $XY$.
8. In Euclidean geometry, the shortest distance from a given point to a straight line is always the perpendicular distance.
9. Therefore, $\mathbf{OP \perp XY}$. (Hence Proved)

Theorem 41: If two tangents are drawn to a circle from an external point, then:

• (i) The lengths of the segments of the tangents from the external point to the points of contact are equal ($PA = PB$).
• (ii) They subtend equal angles at the center ($ngle POA = ngle POB$).
• (iii) The line joining the external point to the center bisects the angle between the tangents ($ngle APO = ngle BPO$).

Given: $P$ is an external point to a circle with center $O$. From $P$, two tangents $PA$ and $PB$ are drawn, touching the circle at points $A$ and $B$ respectively.

To Prove: (i) $PA = PB$, and (ii) $ngle POA = ngle POB$.

Construction: Join $OA$, $OB$, and $OP$.

Proof:

1. By Theorem 40, the tangent at any point of a circle is perpendicular to the radius through the point of contact. $$\implies OA \perp PA \implies ngle OAP = 90^\circ$$ $$\implies OB \perp PB \implies ngle OBP = 90^\circ$$
2. Therefore, $ riangle OAP$ and $ riangle OBP$ are both right-angled triangles.
3. Now, comparing right-angled triangles $ riangle OAP$ and $ riangle OBP$:
   • $ngle OAP = ngle OBP = 90^\circ$ (Right angles)
   • $OP = OP$ (Common hypotenuse)
   • $OA = OB = r$ (Radii of the same circle)
4. By the RHS (Right angle-Hypotenuse-Side) Congruence Criterion: $$ riangle OAP \cong riangle OBP$$
5. Since corresponding parts of congruent triangles (CPCTC) are equal: $$\mathbf{PA = PB} \quad ext{[(i) Lengths of tangent segments are equal]}$$ $$\mathbf{ngle POA = ngle POB} \quad ext{[(ii) Subtend equal angles at center]}$$ $$\mathbf{ngle APO = ngle BPO} \quad ext{[(iii) Center line bisects angle between tangents]}$$

Two circles in a plane are said to touch each other if they share exactly one common point. This common point is called the common point of contact ($T$). There are two distinct geometric configurations:

• External Contact (বহিস্পর্শ): The two circles lie on opposite sides of their common tangent at the contact point. Neither circle lies inside the other. $$\mathbf{d = C_1 C_2 = r_1 + r_2}$$
• Internal Contact (অন্তঃস্পর্শ): One circle lies inside the other, and they share a single common point of contact. Both circles lie on the same side of the common tangent. $$\mathbf{d = C_1 C_2 = |r_1 - r_2|}$$

Theorem: If two circles touch each other (either externally or internally), the point of contact lies on the straight line joining the centers of the two circles (দুটি বৃত্ত পরস্পরকে স্পর্শ করলে স্পর্শবিন্দুটি কেন্দ্রদ্বয়ের সংযোজক সরলরেখার উপর অবস্থিত হবে)।

Proof Sketch: Let circles with centers $C_1$ and $C_2$ touch at point $T$. Let line $XY$ be the common tangent passing through $T$. By Theorem 40, $C_1 T \perp XY$ and $C_2 T \perp XY$. At point $T$ on the straight line $XY$, only one perpendicular line can be drawn in the plane. Hence, the segments $C_1 T$ and $C_2 T$ are collinear, meaning $C_1, T, C_2$ lie on a single straight line.

A line touching two circles simultaneously is called a Common Tangent:

• Direct Common Tangent (DCT - সরল সাধারণ স্পর্শক): Both circles lie on the same side of the tangent line. Length: $$L_{ ext{DCT}} = \sqrt{d^2 - (r_1 - r_2)^2}$$
• Transverse Common Tangent (TCT - তির্যক সাধারণ স্পর্শক): The two circles lie on opposite sides of the tangent line (the line intersects the segment joining centers). Length: $$L_{ ext{TCT}} = \sqrt{d^2 - (r_1 + r_2)^2}$$

Rider: If a quadrilateral $ABCD$ is circumscribed about a circle (all four sides touch the circle), prove that the sum of opposite sides is equal: $$\mathbf{AB + CD = BC + DA}$$

Proof: Let sides $AB, BC, CD, DA$ touch the incircle at points $P, Q, R, S$ respectively. By Theorem 41, the two tangent segments from each vertex to the circle are equal in length:

• From vertex $A$: $AP = AS$
• From vertex $B$: $BP = BQ$
• From vertex $C$: $CR = CQ$
• From vertex $D$: $DR = DS$

In a right-angled triangle with perpendicular sides $a$ and $b$ and hypotenuse $c$, the inradius $r$ of the inscribed circle is given by: $$\mathbf{r = rac{a + b - c}{2} = rac{ ext{Perpendicular} + ext{Base} - ext{Hypotenuse}}{2}}$$ This provides an instantaneous algebraic check for board questions involving right triangles.