The coordinates of the three vertices of a triangle are: (1,2),(7,2),(1,10). Then the radius of the incircle of the triangle is:

The correct option is 2 (which represents a radius of 2).

Let the vertices of the triangle be:
$A=(1,2)$
$B=(7,2)$
$C=(1,10)$

First, let's calculate the lengths of the sides of the triangle using the distance formula. Let:
- $c$ be the length of side $AB$
- $a$ be the length of side $BC$
- $b$ be the length of side $CA$

Calculating side $c$ (distance between $A(1,2)$ and $B(7,2)$):
$c=\sqrt{{(7-1)}^2+{(2-2)}^2}=\sqrt{6^2+0}=6$

Calculating side $b$ (distance between $A(1,2)$ and $C(1,10)$):
$b=\sqrt{{(1-1)}^2+{(10-2)}^2}=\sqrt{0+8^2}=8$

Calculating side $a$ (distance between $B(7,2)$ and $C(1,10)$):
$a=\sqrt{{(1-7)}^2+{(10-2)}^2}=\sqrt{{(-6)}^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$

Notice that the side lengths are $6$, $8$, and $10$. Since $6^2+8^2=36+64=100={10}^2$, the triangle is a right-angled triangle with the right angle at vertex $A$.

The area ($\Delta$) of this right-angled triangle is:
$\Delta =\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 6\times 8=24$

The semi-perimeter ($s$) of the triangle is:
$s=\frac{a+b+c}{2}=\frac{10+8+6}{2}=\frac{24}{2}=12$

The radius of the incircle ($r$) is given by the formula:
$r=\frac{\Delta }{s}$

Substituting the values of $\Delta$ and $s$:
$r=\frac{24}{12}=2$

Thus, the radius of the incircle of the triangle is 2.