Without calculus:
If the circle touches the $x$-axis, then its radius is equal to the absolute value of the $y$-coordinate of its center. So, if $(a,b)$ are the coordinates of its center, an equation of such a circle is $$(x-a)^2+(y-b)^2=b^2.\tag1$$
“Touching” here means that the circle and parabola have a common tangent at their intersection point. If you have a conic with general equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, you can obtain the equation of its tangent at the point $(x_0,y_0)$ via a set of substitutions: $$x^2\to xx_0 \\ xy \to \frac12(xx_0+yy_0) \\ y^2\to yy_0 \\ x\to\frac12(x+x_0) \\ y\to\frac12(y+y_0).$$ So, an equation of the tangent to the parabola at $(2,2)$ is $$x-\frac12(y+2)=0\tag2$$ and an equation of the tangent to the circle is $$2x-a(x+2)+2y-b(y+2)+a^2 = 0.\tag3$$ For these equations to represent the same line, the coefficients of one must be multiples of the corresponding coefficients of the other. Rearranging equations (2) and (3) and comparing coefficients generates the system $$2-a=k \\ 2-b=-\frac12k \\ a^2-2a-2b=-k,$$ which you should be able to solve for $a$ and $b$.
Alternatively, the center of the circle must lie on the perpendicular to the tangent at $(2,2)$. From equation (2), the slope of the tangent at this point is $2$, so an equation of the perpendicular to this tangent is $${y-2\over x-2}=-\frac12.\tag4$$ Additionally, given two tangents to a circle, the circle’s center lies on one of the angle bisectors of those tangents. An equation of the $x$-axis is $y=0$, so the two bisectors of your circle have equations $$\frac1{\sqrt5}(2x-y-2)=\pm y.\tag5$$ Compute the intersections of line (4) with each of the lines in (5).
Another way is similar to how you started to solve the problem. Since $(2,2)$ satisfies $(1)$, we must have $(2-a)^2+(2-b)^2=b^2$. The coordinates of the center must also satisfy equation (4). Solve the resulting system of two equations in $a$ and $b$.
With any of these methods, you will end up with two potential solutions. Reject the solution that doesn’t lie in $D$.