A point M(a; b) lies on the line y = x + 1. M is the center of a circle that passes through K(5; 7) and R(6; 0). Find the coordinates of M and the radius of the circle.

M$(a; b)$ is a point on the line $y = x + 1$. To write $b$ in terms of $a$, substitute the coordinates of M into the equation of the line.

$$b = a + 1$$

The expression for $b$ in terms of $a$ is $\boxed{b = a + 1}$.

To calculate the coordinates of M$(a; b)$, we use the fact that M is the center of the circle and K$(5; 7)$ and R$(6; 0)$ are points on the circle. This means the distance from M to K is equal to the distance from M to R (both are the radius, $r$).

$$MK^2 = MR^2$$ $$(a-5)^2 + (b-7)^2 = (a-6)^2 + (b-0)^2$$

Expand both sides:

$$(a^2 - 10a + 25) + (b^2 - 14b + 49) = (a^2 - 12a + 36) + b^2$$

Subtract $a^2$ and $b^2$ from both sides:

$$-10a - 14b + 74 = -12a + 36$$

Rearrange the terms:

$$2a - 14b + 38 = 0$$

Divide by 2:

$$a - 7b + 19 = 0 \quad (*)$$

Substitute $b = a + 1$ (from Step 1) into equation $(*)$:

$$a - 7(a + 1) + 19 = 0$$ $$a - 7a - 7 + 19 = 0$$ $$-6a + 12 = 0$$ $$-6a = -12$$ $$a = 2$$

Now substitute $a = 2$ back into $b = a + 1$:

$$b = 2 + 1$$ $$b = 3$$

The coordinates of M are $\boxed{(2; 3)}$.

Given that the coordinates of M are $(2; 3)$, we need to calculate the length of the radius of the circle. K$(5; 7)$ is a point on the circle. The radius $r$ is the distance MK.
Using the distance formula:

$$r = \sqrt{(x_K - x_M)^2 + (y_K - y_M)^2}$$ $$r = \sqrt{(5 - 2)^2 + (7 - 3)^2}$$ $$r = \sqrt{(3)^2 + (4)^2}$$ $$r = \sqrt{9 + 16}$$ $$r = \sqrt{25}$$ $$r = 5$$

The radius of the circle is $\boxed{5 \text{ units}}$.

Determine the equation of the tangent to the circle at K$(5; 7)$, given M$(2; 3)$ is the center.
First, find the gradient of the radius MK:

$$m_{MK} = \frac{y_K - y_M}{x_K - x_M}$$ $$m_{MK} = \frac{7 - 3}{5 - 2}$$ $$m_{MK} = \frac{4}{3}$$

The tangent line at K is perpendicular to the radius MK. The gradient of the tangent line, $m_{tangent}$, is the negative reciprocal of $m_{MK}$.

$$m_{tangent} = -\frac{1}{m_{MK}}$$ $$m_{tangent} = -\frac{1}{\frac{4}{3}}$$ $$m_{tangent} = -\frac{3}{4}$$

Now, use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with point K$(5; 7)$ and $m_{tangent} = -\frac{3}{4}$.

$$y - 7 = -\frac{3}{4}(x - 5)$$ $$y - 7 = -\frac{3}{4}x + \frac{15}{4}$$

Add 7 to both sides:

$$y = -\frac{3}{4}x + \frac{15}{4} + 7$$

Convert 7 to a fraction with denominator 4: $7 = \frac{28}{4}$.

$$y = -\frac{3}{4}x + \frac{15}{4} + \frac{28}{4}$$ $$y = -\frac{3}{4}x + \frac{43}{4}$$

The equation of the tangent is $\boxed{y = -\frac{3}{4}x + \frac{43}{4}}$.