A line and a circle have exactly one common point when the line is tangent to the circle. Set the distance from the circle's center to the line equal to the radius.

After determining the allowed integer value of $k$, shift the coordinates using $X=x-1$ and $Y=y+2$. Then substitute the line equation into $X^2+Y^2=4$.

Algebra is faster for finding $k$, but Desmos can verify it. Enter $d(k)=\frac{|5k-10|}{\sqrt{k^2+9}}$ and graph $y=2$. The intersections show the values of $k$ for which the line is tangent to the circle. Choose the positive intersection that is an integer.

Enter $(x-1)^2+(y+2)^2=4$ and $4x+3y=-12$. Click their single intersection to view the coordinates, then add the displayed $x$- and $y$-coordinates to verify the result.

The circle has center $(1,-2)$ and radius $2$. Because the system has exactly one solution, the line must be tangent to the circle.

Rewrite the line as $kx+3y+4k-4=0$. The distance from the center to this line must equal the radius:

Thus,

The possible values are $4$ and $\frac{16}{21}$. Since $k$ is a positive integer, $k=4$.

Substitute $k=4$ into the line equation:

Let $X=x-1$ and $Y=y+2$. Then the circle becomes $X^2+Y^2=4$, and the line becomes $4X+3Y=-10$.

Substitute $Y=\frac{-10-4X}{3}$ into the circle equation:

So $X=-\frac{8}{5}$ and $Y=-\frac{6}{5}$.

Because $X=x-1$ and $Y=y+2$,

Therefore,