Imagine a point outside a circle. Suppose you draw a line from the point passing through the center of the circle, the line intersects the circle at two points. The segment of the line within the area of the circle is equal to the diameter of the circle. When you keep increasing the angle of direction of the line, the points of intersections get closer and closer. The line segment within the circle changes to chords of shorter and shorter lengths from the size of the diameter. Ultimately, the intersection points on the circle coincides each other and the line just touches the circle at only one point. At this condition, the line is called as the ‘tangent’ of the circle from that point.
With further increase in angle of direction of the line will not at all intersect the circle.
In this article let us study the properties of the tangent of a circle.

We have explained in the introduction how a line from a point outside the circle could be a tangent to that circle. Now let us formally define what a tangent of a circle is. A line drawn from a point outside a circle which touches the circle just at one point is called tangent to the circle. The point of contact of the tangent and the circle is called as the ‘point of tangency’.
Tangent of a Circle
The above diagram shows a circle with center O and a tangent PA drawn from a point P outside the circle. The point A is called the point of tangency.
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.

Tangent of a Circle Formula
In the above diagram, as per this theorem m of < PAO = 90o.
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 = 90o (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 = 90o
Tangents drawn from the same point to a circle are equal in measure.
Finding Tangent of a Circle
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 = 90o.
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 PB
The 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.
Find Tangent of a 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 = 90o - < PAB
In triangle OAB,
< AOB + < OAB + < OBA = 180o.
or, < AOB + 2< OAB = 180o ( since < OAB = < OBA)
or, < AOB + 2(90o - < PAB) = 180o
or, < AOB + 180o - 2< PAB = 180o
So, < AOB = 2< PAB
Now, < ACB = ($\frac{1}{2}$) < AOB (Angle at the center is twice the angle at circumference)
Or, < ACB = < PAB
We will give examples for finding tangent of a circle by way of some illustrated problems.

Solved Examples

Question 1: Two tangents are drawn from a point 12 in. from the center of a circle which has a diameter of 10 in. Find the length of each tangent.

How to Find Tangents of a Circle
Solution:
 
Let P be the point from which the tangents PA and PB are drawn to the circle with center O. As per the properties of tangent of a circle, < PAO = < PBO = 90o and also PA = PB.
It is given that OA = OB = $\frac{10in}{2}$ = 5 in. and PO = 12 in.
The triangle PAO is right triangle.
Therefore, PA2 = PO2 + OA2
or, PA2 = (12)2 + (5)2
or, PA2 = 144 + 25 = 169
or, PA = 13 in. As explained already PB also = 13 in.
Thus, the length of the tangents is 13in.
 

Question 2: In the following diagram, < AOB = 120o and OA = 15cm. Find the length of the tangent AP.
Tangent of a Circle Problems
Solution:
 
In triangle AOB, the angle AOB = 120o.

Therefore, the angles AOD and BOD are each equal to 60o.
Also, < OAD = < OBD = $\frac{(180^o - 120^o)}{2}$ = 30o
Now AD = OA cos OAD = 15 $\times$ cos 30o ≈  15 $\times$ (0.866) cm  = 13 cm
As per properties of circle, BD = AD = 13 cm
AB = AD + BD = 2 AD = 26 cm.
Now angle ACB = ($\frac{1}{2}$) < AOB
                        = ($\frac{1}{2}$)( 120o) = 60o. (Angle at the center is twice the angle at circumference)
So, < BAP = 60o. (Alternate segment theorem because PA is a tangent)
AP = AB cos BAP = AB cos 60o= 26 $\times$ ($\frac{1}{2}$) = 13 cm.
Thus the length of the tangent AP = 13 cm.