The least distance of the line $8x-4y+73=0$ from the circle $16x^2+16y^2+48x-8y-43=0$ is

$x^2+y^2+3x-\frac{y}{2}-\frac{43}{16}=0$

$\text{centre}(-\frac{3}{2},\frac{1}{4}),CA=r=\sqrt{\frac{9}{4}+\frac{1}{16}+\frac{43}{16}}=\sqrt{5}$

$CP=\left|\frac{8(-\frac{3}{2})-4\left(\frac{1}{4}\right)+73}{\sqrt{64+16}}\right|=3\sqrt{5}$

$AP=CP-CA=3\sqrt{5}-\sqrt{5}=2\sqrt{5}$