If a circle and a line are drawn on a paper, three things can possible happen:

(i) The line does not touches or cuts the circle. As you see in the figure besides that $XY$ does not touch or cut the circle.

(ii) The line cuts the circle in two parts. The straight line which cuts the circle in two points is called the secant of the circle. In this case the line $PS$ cuts the circle at two points $Q$ and $R$. $QR$ is a chord.

(iii) The line touches the circle at only one point. The line that touches the circle at only one point is called tangent of the circle. The point at which the line touches the circle is called the point of contact. In this case, you see that the line $AB$ touches the circle at point $C$ (one point only).

Theorems related to tangents:

Theorem 18: The tangent at any point of a circle and the radius through this point are perpendicular to each other.

Given: $AB$ is a tangent. Point of contact is $P$.

To prove: $OP \perp AB$

Proof: $OP < OQ$ (since $Q$ is outside the circle)

Similarly, we can show that out of all possible line segments that could be drawn from $O$ to the line $AB, OP$ would be the shortest. Hence $OP \perp AB$.  (Reason: The shortest line segment, drawn from a given point to a given line is perpendicular to this line)

Hence Proved

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

(i) the lengths of the tangents to the circle are equal

(ii) the tangents will subtend equal angles at the center of the circle

(iii) tangents are equally inclined to the line joining the point and the center of the circle.

Given: A circle with center $O$. $PA$ and $PB$ are two tangents to this circle from external point $P$.

To Prove:

(i) $PA = PB$

(ii) $\angle AOP = \angle BOP$

(iii) $\angle APO = \angle BPO$

Proof:

Consider $\triangle AOP$ and $\triangle BOP$

$OA = OB$ (radius of the same circle)

$\angle OAP = \angle OBP = 90^o$ (Theorem 18)

$OP$ is common

$\therefore \triangle AOP \cong \triangle BOP$ (by R.H.S postulate)

Since corresponding parts of the congruent triangles are equal, we get

(i) $PA = PB$

(ii) $\angle AOP = \angle BOP$

(iii) $\angle APO = \angle BPO$

Hence Proved

Theorem 19: If two circles touch each other, the point of contact lines on the straight line through the centers.

There are two possible scenarios

Case 1: when the circles just touch each other externally

Given: Two circles with centers $A$ and $B$ touch each other externally at point $P$ as shown in the diagram.

To Prove: $P$ lies on the line $AB$ i.e. $A, P$ and $B$ are collinear.

Proof:

$\angle APQ = 90^o$ (angle between the radius and the tangent)

$\angle BPQ = 90^o$ (angle between the radius and the tangent)

$\angle APQ \angle BPQ = 180^o$

$\Rightarrow \angle APB = 180^o$ which means that $APB$ is a straight line.

Hence Proved

Case 2: when the two circles touch each other internally

Given: Two circles with centers $A$ and $B$ touch each other internally at point $P$ as shown in the diagram.

To Prove: $P$ lies on the line $AB$ i.e. $A, B$ and $P$ are collinear.

Proof:

$\angle APQ = 90^o$ (angle between the radius and the tangent)

$\angle BPQ = 90^o$ (angle between the radius and the tangent)

Therefore both $AP$ and $BP$ are perpendicular to the tangent $PQ$ at the point $P$.

Therefore $AP$ and $BP$ lie on the same line because only one perpendicular can be drawn through a line through a point on it.

Hence Proved

Theorems related to Chords:

Theorem 20: If two chords of a circle intersect internally or externally then the product of the lengths of their segments is equal.

There are two possible cases.

Case 1: When the chords intersect internally

Given: Chords $AB$ and $CD$ of a circle intersect each other at point $P$ inside the circle.

To Prove: $PA \times PB = PC \times PD$

Proof: Consider $\triangle APC$ and $\triangle PBD$

$\angle CAB = \angle CDB$ (angles in the same segment)

$\angle ACD = \angle DBA$ (angles in the same segment)

Therefore $\triangle APC \sim \triangle PBD$ (AAA Postulate)

$\Rightarrow \frac{PA}{PD} =\frac{PC}{PB}$ (corresponding sides of similar triangles are proportional)

$\Rightarrow PA \times PB = PC \times PD$

Hence Proved

Case 2: When the chords intersects externally

Given: Chords $AB$ and $CD$ of a circle, when produced,  intersect each other at point $P$ outside the circle.

To Prove: $PA \times PB = PC \times PD$

Proof: Consider $\triangle PAC$ and $\triangle PDB$

$\angle PAC = \angle PDB$ (external angle of a cyclic quadrilateral is equal to internal opposite angle)

$\angle ACP = \angle PBD$ (external angle of a cyclic quadrilateral is equal to internal opposite angle)

$\triangle PAC \sim \triangle PDB$ (By AAA postulate)

$\Rightarrow \frac{PA}{PD} =\frac{PC}{PB}$ (corresponding sides of similar triangles are proportional)

$\Rightarrow PA \times PB = PC \times PD$

Hence Proved

Theorem 21: The angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.

Given: A circle with center $O$. Tangent $PQ$ touches the circle at point $A$. $AB$ is a chord drawn through point $A$.

To Prove: $\angle BAQ = \angle ACB$ and $\angle BAP = \angle ADB$

Proof: In $\triangle ABR$

$\angle ABR = 90^o$ (angle in a semi circle)

$\Rightarrow \angle ARB + \angle RAB = 90^o$

$\angle OAQ = 90^o$ (angle between the radius and the tangent)

$\Rightarrow \angle RAB + \angle BAQ = 90^o$

$\angle ARB + \angle RAB = \angle RAB + \angle BAQ$

$\Rightarrow \angle ARB = \angle BAQ$

But $\angle ARB = \angle ACB$ (angles in the same segment)

$\therefore \angle BAQ = \angle ACB$

Now $\angle BAP + \angle BAQ = 180^o$ ($PAQ$ is a straight line)

$\angle ACB + \angle ADB = 180^o$ (Opposite angles of a cyclic quadrilateral)

$\Rightarrow \angle BAP + \angle BAQ = \angle ACB + \angle ADB$

$\Rightarrow \angle BAP = \angle ADB$ as $\angle BAQ = \angle ACB$

Hence Proved

Theorem 22: If a chord and a tangent intersect externally, then the product of the length of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

Given: Chord $AB$ and tangent $TP$ of a circle intersect each other at point P outside the circle.

To Prove: $PA \times PB = PT^2$

Proof: Consider $\triangle PAT$ and $\triangle PTB$

$\angle PTB = \angle TAB$ (Angles in alternate segment)

$\angle TPA = \angle TPB$ (common angle)

$\triangle PAT \sim \triangle PTB$ (AAA Postulate)

$\Rightarrow \frac{PA}{PT} =\frac{PT}{PB}$ (corresponding sides of similar triangles are proportional)

$\Rightarrow PA \times PB = PT^2$

Hence Proved