Solve problems involving coordinate geometry and circle geometry, including finding the equation of a line, distance between points, and equation of a circle.

Given points are $(x_1, y_1) = (1,2)$ and $(x_2, y_2) = (3,6)$.
$$
m = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2
$$

Using point $(1,2)$ and slope $m=2$:
$$
y - 2 = 2(x - 1)
$$
$$
y - 2 = 2x - 2
$$
$$
y = 2x - 2 + 2
$$
$$
y = 2x
$$
The equation of the line is $\boxed{y = 2x}$.

Given points are $(x_1, y_1) = (2,3)$ and $(x_2, y_2) = (6,7)$.
$$
D = \sqrt{(6 - 2)^2 + (7 - 3)^2}
$$
$$
D = \sqrt{(4)^2 + (4)^2}
$$
$$
D = \sqrt{16 + 16}
$$
$$
D = \sqrt{32}
$$
$$
D = \sqrt{16 \times 2}
$$
$$
D = 4\sqrt{2}
$$
The distance between the points is $\boxed{4\sqrt{2} \text{ units}}$.

Given center $(h,k) = (0,0)$ and radius $r=5$.  
Substitute these values into the equation:
$$
(x-0)^2 + (y-0)^2 = 5^2
$$
$$
x^2 + y^2 = 25
$$
The equation of the circle is $\boxed{x^2 + y^2 = 25}$.

Substitute $x_1=3$, $y_1=4$, and $r^2=25$:
$$
x(3) + y(4) = 25
$$
$$
3x + 4y = 25
$$
The equation of the tangent is $\boxed{3x + 4y = 25}$.