It is given that $x = -2$ and $y = -1$ are tangents to a circle.
The $x$-coordinate and $y$-coordinate of the centre of the circle are positive.
The line $3y = 2x + 5$ is a normal to the circle.

1. Express the radius of the circle in terms of the coordinates of its centre.

Let the centre of the circle be $(h, k)$ and its radius be $r$.
Since $x = -2$ is a tangent and $h > 0$, the distance from the centre to this tangent is $r = h - (-2) = h + 2$.
Since $y = -1$ is a tangent and $k > 0$, the distance from the centre to this tangent is $r = k - (-1) = k + 1$.

2. Form an equation relating the coordinates of the centre.

Equating the two expressions for the radius:
$h + 2 = k + 1$
$k = h + 1$

3. Use the property of the normal to form a second equation.

A normal to a circle always passes through its centre. Therefore, the centre $(h, k)$ lies on the line $3y = 2x + 5$.
Substituting $x = h$ and $y = k$ into the equation:
$3k = 2h + 5$

4. Solve the simultaneous equations to find the centre.

Substitute $k = h + 1$ into $3k = 2h + 5$:
$3(h + 1) = 2h + 5$
$3h + 3 = 2h + 5$
$h = 2$
Substitute $h = 2$ back into $k = h + 1$:
$k = 2 + 1 = 3$
Since $h = 2 > 0$ and $k = 3 > 0$, this satisfies the given conditions.
Thus, the centre of the circle is $(2, 3)$ (shown).

1. Find the radius of the circle

The radius is the perpendicular distance from the centre $(2, 3)$ to the tangent $x = -2$ (or $y = -1$).
$r = 2 - (-2) = 4$

2. Form the equation of the circle

Using the standard form $(x - a)^2 + (y - b)^2 = r^2$:
$(x - 2)^2 + (y - 3)^2 = 4^2$
$(x - 2)^2 + (y - 3)^2 = 16$