### In the given figure, $AB$ is a chord of length $13 \space cm$ of a circle with center $O$ and radius $9 \space cm$. The tangents at $A$ and $B$ intersect at $P$. Find the length of $PA$. O B A P

$9.4 \space cm$

Step by Step Explanation:
1. Given:
Length of chord $AB$ is $13 \space cm$.
The radius of the circle is $9 cm$.
The tangents at $A$ and $B$ intersect at $P$.
2. Here, we have to find the length of $PA$.

Now, join $O$ to $P$ such that $OP$ intersects $AB$ at $M$.

Let $PA = x \space cm$ and $PM = y \space cm$.
3. The tangents from an external point are equal in length. $$\implies PA = PB$$ Also, two tangents to a circle from an external point are equally inclined to the line segment joining the center to that point. \begin{aligned} \implies & OP \text{ is the bisector of } \angle APB. \\ \implies & \angle APM = \angle MPB \end{aligned} Also, \begin{aligned} PO = PO && \text{[Common]} \end{aligned} By SAS Congruence Criterion, we conclude \begin{aligned} \triangle APM \cong \triangle MPB \end{aligned}
4. As corresponding parts of congruent triangles are equal$(CPCT)$, we have $$\angle PMA = \angle PMB \text{ and } AM = BM$$ Also, \begin{aligned} & \angle PMA + \angle PMB = 180^\circ && \text{ [Angles on a straight line] } \\ \implies & 2 \angle PMA = 180^\circ && \text{ [As, } \angle PMA = \angle PMB] \\ \implies & \angle PMA = \angle PMB = 90^\circ \end{aligned}
5. Now, we can say that $OP$ is the right bisector of $AB$.

Thus, $OP \perp AB$ and $OP$ bisects $AB$ at $M$.

Therefore, $AM = BM = \dfrac { 13 } { 2 } \space cm = 6.5 \space cm$.
6. Now, \begin{aligned} & \angle PMA = 90^\circ \implies \angle AMO = 90^\circ && \text{ [Angles on a straight line.] } \\ \implies & \triangle AMO \text{ is a right-angled triangle.} \end{aligned}

In right $\triangle AMO$, we have \begin{aligned} OA = 9 \space cm && \text{[Radius]} \\ AM = 6.5 \space cm && \text{[From step 5]} \end{aligned} Therefore, using pythagoras theorem, we have \begin{aligned} OM^2 & = AM^2 + OM^2 \\ OM & = \sqrt{(9)^2 - (6.5)^2} \space cm \\ & = \sqrt{ 38.75 } \space cm \\ & = 6.22 \space cm \end{aligned}
7. Also, $\triangle APM$ is a right angled triangle.
Using pythagoras theorem, we have \begin{aligned} & PA^2 = PM^2 + AM^2 \\ \implies & x^2 = y^2 + (6.5)^2 \\ \implies & x^2 = y^2 + 42.25 && \ldots \text{(i)} \end{aligned}
8. In right $\triangle PAO$, using pythagoras theorem \begin{aligned} & OP^2 = PA^2 + OA^2 && [\angle PAO = 90^\circ \text{ as AO is the } \space \\ & && \text { radius at the point of contact.] } \\ \implies & (PM + OM)^2 = x^2 + (9)^2 && \text{[As, } OP = PM + OM ] \\ \implies & PM^2 + OM^2 + 2 \times PM \times OM = x^2 + 81 \\ \implies & y^2 + 38.75 + 2y \times 6.22 = x^2 + 81 \\ \implies & y^2 + 38.75 + 2y \times 6.22 = y^2 + 42.25 + 81 && \text{[By using } eq \text{(i)} ] \\ \implies & 12.44 y = 84.5 \\ \implies & y = 6.79 \space cm \end{aligned}
9. Now, substituting the value of $y$ in $\text{(i)}$, we get \begin{aligned} & x^2 = (6.79)^2 + 42.25 = 46.14 + 42.25 = 88.39 \\ \implies & x = \sqrt { 88.39 } = 9.4 \end{aligned}
Thus, $PA = 9.4 \space cm.$ 