### Theory:

Result $$3$$:
The lengths of the two tangents drawn from an exterior point to a circle are equal.
Explanation:

The lengths of the two tangents $$PA$$ and $$PB$$ drawn from an exterior point $$P$$ to a circle are equal.

$$\Rightarrow$$ $$PA = PB$$.
Proof for the result:
By the result $$1$$, we have:
A tangent at any point on a circle and the radius through the point are perpendicular to each other.
$$OB$$ $$\perp$$ $$PB$$  and $$OA$$ $$\perp$$ $$PA$$.

Here, $$OA$$ and $$OB$$ are radius. Hence, they are equal.

The side $$OP$$ is a common side to the triangles $$AOP$$ and $$BOP$$.

Thus, the triangles $$AOP$$ and $$BOP$$ are congruent.

Therefore, $$PA = PB$$.
Example:
In the above given figure if $$OB$$ $$=$$ $$3$$ $$cm$$  and $$OP$$ $$=$$ $$5$$ $$cm$$, find the length of $$PA$$.

Solution:

By the result $$1$$, we have:
A tangent at any point on a circle and the radius through the point are perpendicular to each other.
$$\angle OPB$$ $$=$$ $$90^{\circ}$$.

So, $$OPB$$ is a right angled triangle.

By the Pythagoras theorem, we have:
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
$$OP^2$$ $$=$$ $$OB^2$$ $$+$$ $$PB^2$$

$$PB^2$$ $$=$$ $$OP^2$$ $$-$$$$OB^2$$

$$PB^2$$ $$=$$ $$5^2$$ $$-$$ $$3^2$$

$$PB^2$$ $$=$$ $$25 - 9$$

$$PB^2$$ $$=$$ $$16$$

$$\Rightarrow$$  $$PB$$ $$=$$ $$\sqrt{16}$$

$$PB$$ $$=$$ $$4$$

Thus, the measure of $$PB$$ $$=$$ $$4$$ $$cm$$

By the result $$3$$, we have:
The lengths of the two tangents drawn from an exterior point to a circle are equal.
Hence, $$PA$$ $$=$$ $$PB$$.

Therefore, the measure of $$PA$$ $$=$$ $$4$$ $$cm$$