Find the equation of the circle, coordinates of C, and the equation of tangent AB.

The center of the circle is $M(-2; 3)$, so $a = -2$ and $b = 3$.
The circle passes through point $A(1; -1)$.
The radius squared, $r^2$, is the distance squared between M and A:

The equation of the circle is:

The center of the circle is $M(-2; 3)$ and the radius is $r = \sqrt{25} = 5$.
The line BC is parallel to the y-axis, meaning it is a vertical line.
Since BC is tangent to the circle at C, the radius MC is perpendicular to BC.
As BC is vertical, MC must be horizontal. This means the y-coordinate of C is the same as the y-coordinate of M, which is 3.
From the diagram, C is to the left of M. The x-coordinate of C is $x_M - r$.

The coordinates of C are:

The tangent AB touches the circle at $A(1; -1)$. The center is $M(-2; 3)$.
First, find the gradient of the radius MA:

Since the tangent AB is perpendicular to the radius MA, the product of their gradients is -1:

Now use the point-slope form $y - y_1 = m(x - x_1)$ with point $A(1; -1)$ and $m_{AB} = \frac{3}{4}$:

The equation of the tangent AB is:

From Step 2, $C(-7; 3)$.
The line BC is a vertical line, so the x-coordinate of B is the same as C, which is -7.
Point B is the intersection of the tangent AB (from Step 3) and the line $x = -7$.
Substitute $x = -7$ into the equation of AB:

So, $B(-7; -7)$.
The length of BC is the absolute difference in their y-coordinates:

The length of BC is:

The center of this new circle is $A(1; -1)$.
For a circle to be tangent to both the x-axis and the y-axis, its radius must be equal to the absolute value of its x-coordinate and y-coordinate.
Radius $r = |x_A| = |1| = 1$.
Radius $r = |y_A| = |-1| = 1$.
So, the radius of this circle is $r = 1$.
The equation of the circle is $(x - x_A)^2 + (y - y_A)^2 = r^2$:

The equation of the circle is:

The original circle (Circle M) has center $M(-2; 3)$ and radius $r_M = 5$ (from Step 1).
The new circle (Circle N) has center $N(p