Two circles, having their centres at A and at B, touch externally at a point M. From a point P on a tangent at M, tangents PQ and PR are drawn to the circles with points of contact Q and R, respectively.

If AP = 13 cm and AM = 5 cm, find the value of $PQ+PR-PM$.

The correct option is 12 cm.

Let's break down the geometric properties and solve the problem step-by-step.
1. First, let's understand the configuration of the circles and the tangents:
- We have two circles touching externally at a point $M$. The center of the first circle is at $A$, and its radius is $AM=5\text{cm}$.
- A common tangent is drawn at the contact point $M$. Let this tangent be $PM$, where $P$ is a point on it. Since $PM$ is tangent to the circle centered at $A$ at the point of contact $M$, the radius $AM$ is perpendicular to the tangent line $PM$. Therefore, $\angle AMP=90°$.
- This makes $\Delta AMP$ a right-angled triangle at $M$.

2. We can use the Pythagorean theorem on $\Delta AMP$ to find the length of $PM$:
$A{P}^2=A{M}^2+P{M}^2$
Given that $AP=13\text{cm}$ and $AM=5\text{cm}$, we substitute these values:
${13}^2=5^2+P{M}^2$
$169=25+P{M}^2$
$P{M}^2=169-25=144$
$PM=\sqrt{144}=12\text{cm}$.

3. Next, let's look at the tangents drawn from point $P$:
- From an external point $P$, the lengths of tangents drawn to a circle are equal.
- For the first circle (centered at $A$), the tangents drawn from $P$ are $PQ$ and $PM$. Therefore, $PQ=PM$.
- For the second circle (centered at $B$), the tangents drawn from $P$ are $PR$ and $PM$. Therefore, $PR=PM$.
- Consequently, we have the equality: $PQ=PR=PM=12\text{cm}$.

4. Finally, we calculate the required expression:
$PQ+PR-PM=12+12-12=12\text{cm}$.