Determine the centre and radius of the second circle.

The equation of the second circle is given by $x^2 + y^2 + x - 3y + 2 = 0$.
Comparing this to the general form of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, we have:
$2g = 1 \implies g = \frac{1}{2}$
$2f = -3 \implies f = -\frac{3}{2}$
$c = 2$

The coordinates of the centre are $(-g, -f)$.
Centre $= \left(-\frac{1}{2}, -\left(-\frac{3}{2}\right)\right) = \left(-\frac{1}{2}, \frac{3}{2}\right)$.

The radius $R$ is given by $\sqrt{g^2 + f^2 - c}$.
$R = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{3}{2}\right)^2 - 2}$
$R = \sqrt{\frac{1}{4} + \frac{9}{4} - 2}$
$R = \sqrt{\frac{10}{4} - \frac{8}{4}}$
$R = \sqrt{\frac{2}{4}}$
$R = \sqrt{\frac{1}{2}}$
$R = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

The centre of the circle is $\boxed{\left(-\frac{1}{2}, \frac{3}{2}\right)}$ and the radius is $\boxed{\frac{\sqrt{2}}{2}}$.

The common chord AB has the equation $y = -x + 1$.
Point A is an intersection of this line and the second circle $x^2 + y^2 + x - 3y + 2 = 0$.
Substitute $y = -x + 1$ into the circle equation:

This gives two possible x-coordinates: $x = 0$ or $x = -1$.

If $x = 0$, then $y = -0 + 1 = 1$. This gives the point $(0, 1)$.
If $x = -1$, then $y = -(-1) + 1 = 1 + 1 = 2$. This gives the point $(-1, 2)$.

From the problem description, the first circle touches the y-axis at point B. The point $(0,1)$ lies on the y-axis. Therefore, B is $(0,1)$.
Since A and B are the intersection points, A must be the other point.
The coordinates of A are $\boxed{(-1, 2)}$.

The first circle is centred at M and touches the x-axis at C and the y-axis at B.
From Step 2, we found B is $(0,1)$.
Since the circle touches the y-axis at $(0,1)$, its radius must be $r=1$.
Also, the y-coordinate of the centre M must be 1.
Since the circle touches the x-axis at C, and from the diagram C is on the negative x-axis, the x-coordinate of the centre M must be $-r = -1$.
So, the centre M is $(-1, 1)$ and the radius is $r=1$.

The equation of a circle with centre $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting M $(-1, 1)$ and $r=1$:

This matches the required equation.

The straight line $y = -x + k$ is a tangent to the circle with centre M, which is $x^2 + y^