Case 1 : There is no tangent to a circle passing through a point lying inside the circle.

Case 2 : There is one and only one tangent to a circle passing through a point lying on the circle.

Case 3 : There are exactly two tangents to a circle through a point lying outside the circle.

The length of the segment of the tangent from the external point P and the point of contact with the circle is called the length of the tangent from the point P to the circle.

Proof : Given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P. Prove that PQ = PR.

For this, join OP, OQ and OR.
Then ∠ OQP and ∠ ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles.
Now in right triangles OQP and ORP,

OQ = OR (Radii of the same circle)

OP = OP (Common)

Therefore, ∆ OQP ≅ ∆ ORP ........(RHS)

This gives PQ = PR ........ (CPCT)

Remarks :
1. The theorem can also be proved by using the Pythagoras Theorem as follows: $PQ^2$ = $OP^2 – OQ^2 $ = $OP^2 – OR^2$ = $PR^2$ (As OQ = OR) which gives PQ = PR.

2. Note also that ∠ OPQ = ∠ OPR. Therefore, OP is the angle bisector of ∠ QPR, i.e., the centre lies on the bisector of the angle between the two tangents.