# Length Of Tangent On A Circle

A tangent to a circle is defined as a line segment that touches the circle exactly at one point. There are some important points regarding tangents:

• A tangent to a circle cannot be drawn through a point which lies inside the circle. It is so because all the lines passing through any point inside the circle, will intersect the circle at two points.
• There is exactly one tangent to a circle which passes through only one point on the circle.
• There are exactly two tangents can be drawn to a circle from a point outside the circle.

In the figure, $P$ is an external point from which tangents are drawn to the circle. $A$ and $B$ are points of contact of the tangent with a circle. The length of a tangent is equal to the length of a line segment with end-points as the external point and the point of contact. So, $PA$ and $PB$ are the lengths of tangent to the circle from an external point $P$.

## Some theorems on length of tangent

Theorem 1: The lengths of tangents drawn from an external point to a circle are equal. It is proved as follows:

Consider the circle with center $O$. $PA$ and $PB$ are the two tangents drawn to the circle from the external point $P$. $OA$ and $OB$ are radii of the circle.

Since tangent on a circle and the radius are perpendicular to each other at the point of tangency,

$∠PAO$ = $∠PBO$ = $90°$

Consider the triangles, $∆PAO$ and $∆PBO$,

$∠PAO$ = $∠PBO$ = $90°$

$PO$ is common side for both the triangles,

$OA$ = $OB$ [Radii of the circle]

Hence, by RHS congruence theorem,

$∆PAO ≅ ∆PBO$

$⇒ PA = PB$ (Corresponding parts of congruent triangles)

This can also be proved by using Pythagoras theorem as follows,

Since,

$∠PAO$ = $∠PBO$ = $90°$

$∆PAO$ and $∆PBO$ are right angled triangles.

$PA^2$ = $OP^2 – OA^2$

Since $OA$ = $OB$,

$PA^2$ = $OP^2 – OB^2$ = $PB^2$

This gives, $PA$ = $PB$

Therefore, tangents drawn to a circle from an external point will have equal lengths. There is an important observation here:

• Since $∠APO$ = $∠BPO$, $OP$ is the angle bisector of $∠APB$.

Therefore, center of the circle lies on angle bisector of the angle made by two tangents to the circle from an external point.

Let’s consider an example for better understanding of the concept of length of the tangent drawn to a circle from an external point.

Example: A circle is inscribed in the quadrilateral $ABCD$, Prove that $AB + CD$ = $AD + BC$.

Tangents drawn from the point $A$ will have equal lengths.

This gives,

$AP$ = $AM$                                —(1)

Similarly, for tangents drawn from point $B$,

$BN$ = $BM$                              —(2)

From point $C$,

$CN$ = $CO$                             —(3)

From point $D$,

$DP$ = $DO$                             —(4)

Adding equations(1),(2), (3) and (4) gives,

$AP + BN + CN + DP$ = $AM + BM + CO + DO$

$\Rightarrow AP + PD + BN + NC$ = $AM + MB + DO + OC$

$\Rightarrow AD + BC$ = $AB + CD$

That’s the required proof.