The equation of the tangent to the circle $x^2+y^2+2gx+2fy+c=0$ at a point $\left(x_1, y_1\right)$ is $x{x}_1+y{y}_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0.$

Thus, the equation of the tangent to $x^2+y^2+4x-6y-12=0$at $\left(1, -1\right)$ is

$x\left(1\right)+y\left(-1\right)+2\left(x+1\right)-3\left(y-1\right)-12=0$

$\Rightarrow 3x-4y-7=0.$

Also, we know that the equation of a circle touching a circle $x^2+y^2+2gx+2fy+c=0$ and line $lx+my+n=0$ at  $\left(x_1, y_1\right)$ is ${\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2+\lambda \left(lx+my+n\right)=0$

Hence, the required circle touching the line $3x-4y-7=0$ at $\left(1, -1\right)$ is

${\left(x-1\right)}^2+{\left(y+1\right)}^2+\lambda \left(3x-4y-7\right)=0$

Given, this circle passes through $\left(4, 0\right)$

$\Rightarrow 9+1+\lambda \left(12-7\right)=0$

$\Rightarrow \lambda =-2$

Hence, the equation of the circle is ${\left(x-1\right)}^2+{\left(y+1\right)}^2-2\left(3x-4y-7\right)=0$

$\Rightarrow x^2+y^2-8x+10y+16=0$

Also, we know that the radius of the circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{g^2+f^2-c}$

Hence, radius $=\sqrt{{\left(-4\right)}^2+5^2-16}=\sqrt{16+25-16}=5$ units.