There are a number of theorems on the tangent of a circle. We will study a few of them which are important. The theorems are in the form of statements. We will describe each and give an algebraic interpretation of the same with the help of diagrams. The algebraic interpretation is nothing but a tangent of a circle formula.

A tangent of a circle makes a right angle to the radius of the circle at the point of tangency.

In the above diagram, as per this theorem m of < PAO = 90^{o}.This theorem can be proved by reductio ad absurdum (proof by contradiction).

Suppose that PAO is not a right angle. Then, let OB be the perpendicular drawn from the center of the circle to the tangent PC. Hence, m < PBO is a right angle.

Now in triangle ABO,

angle ABO = 90

^{o} (as OB is perpendicular to PC)

Hence, triangle ABO is right triangle.

So, OA = hypotenuse of the right triangle (side opposite to the right angle)

And, OB = leg of the right triangle (adjacent side to the right angle)

Therefore, OA > OB (hypotenuse is always longer than the leg)

Or, OA > OA’ + A’B

OA > OA + A’B (since OA’ = OA, both are radius of the same circle)

This is an impossible statement unless A’B is 0.

Hence the assumption that OB is perpendicular to the tangent PC is false. Same way we can show OC also cannot be a perpendicular, where ‘C’ is a point on the other side of A. Since A and C are arbitrary points on the tangent, the line to any point other than A on the tangent cannot be a perpendicular and only OA can be the perpendicular.

Hence, m of < PAO = 90^{o} Tangents drawn from the same point to a circle are equal in measure.

In the above diagram, as per this theorem m of PA = m of PBIt can be logically concluded that only two tangents can be drawn to a circle from the same point outside the circle. One tangent can be on one direction to the line joining the given point and the center of the circle and the other will be on the opposite direction. In the diagram shown, OA and OB are the radii at the points of tangency.

As per radius and tangent property, the angle made by the tangent with the radius at the point of tangency is right angle.

Hence, angle PAO = angle PBO = 90

^{o}.

In right triangles PAO and PBO,

AO = BO (radii of the same circle)

PO = PO (reflexive property)

Hence, triangles PAO and PBO are congruent as per HL congruency.

Therefore, as per CPCT, m of PA = m of PBThe angle made by a tangent with the chord of the circle is congruent to the angle made by the same chord on alternate segment of the circle.

In the above diagram, as per this theorem < PAB = < ACBThe above theorem is proved as follows.

Join AO and PO.

In triangle OAB, OA = OB (radii of the same circle)

Hence triangle OAB is isosceles.

Therefore, < OAB = < OBA

Now angle OAB = 90

^{o} - < PAB

In triangle OAB,

< AOB + < OAB + < OBA = 180

^{o}.

or, < AOB + 2< OAB = 180

^{o} ( since < OAB = < OBA)

or, < AOB + 2(90

^{o} - < PAB) = 180

^{o}or, < AOB + 180

^{o} - 2< PAB = 180

^{o}So, < AOB = 2< PAB

Now, < ACB = (

$\frac{1}{2}$) < AOB (Angle at the center is twice the angle at circumference)

Or, < ACB = < PAB