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

Explanation

To find the least distance of the line \(8x - 4y + 73 = 0\) from the circle given by \(16x^2 + 16y^2 + 48x - 8y - 43 = 0\), we will follow these steps:

### Step 1: Convert the Circle Equation to Standard Form

The equation of the circle is given as:
\[ 16x^2 + 16y^2 + 48x - 8y - 43 = 0 \]
To convert this into standard form, we divide the entire equation by 16:
\[ x^2 + y^2 + 3x - \frac{1}{2}y - \frac{43}{16} = 0 \]

### Step 2: Rearranging the Circle Equation

Rearranging gives:
\[ x^2 + y^2 + 3x - \frac{1}{2}y = \frac{43}{16} \]

### Step 3: Completing the Square

Next, we complete the square for both \(x\) and \(y\):

1. For \(x^2 + 3x\):
\[ x^2 + 3x = \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} \]

2. For \(y^2 - \frac{1}{2}y\):
\[ y^2 - \frac{1}{2}y = \left(y - \frac{1}{4}\right)^2 - \frac{1}{16} \]

Substituting these back into the equation gives:
\[ \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + \left(y - \frac{1}{4}\right)^2 - \frac{1}{16} = \frac{43}{16} \]

### Step 4: Simplifying the Circle Equation

Combining constants:
\[ \left(x + \frac{3}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{43}{16} + \frac{9}{4} + \frac{1}{16} \]
Converting \(\frac{9}{4}\) to sixteenths:
\[ \frac{9}{4} = \frac{36}{16} \]
Thus,
\[ \frac{43}{16} + \frac{36}{16} + \frac{1}{16} = \frac{80}{16} = 5 \]

So, the equation of the circle in standard form is:
\[ \left(x + \frac{3}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = 5 \]
The center of the circle is \(\left(-\frac{3}{2}, \frac{1}{4}\right)\) and the radius is \(\sqrt{5}\).

### Step 5: Finding the Perpendicular Distance from the Center to the Line

The line equation is \(8x - 4y + 73 = 0\). We can use the formula for the distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 8\), \(B = -4\), \(C = 73\), and the center \((x_1, y_1) = \left(-\frac{3}{2}, \frac{1}{4}\right)\).

Calculating:
\[ d = \frac{|8 \left(-\frac{3}{2}\right) - 4 \left(\frac{1}{4}\right) + 73|}{\sqrt{8^2 + (-4)^2}} \]
\[ = \frac{|-12 - 1 + 73|}{\sqrt{64 + 16}} = \frac{|60|}{\sqrt{80}} = \frac{60}{4\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5} \]

### Step 6: Finding the Least Distance from the Circle

The least distance from the line to the circle is the perpendicular distance from the center to the line minus the radius of the circle:
\[ \text{Least Distance} = d - r = 3\sqrt{5} - \sqrt{5} = 2\sqrt{5} \]

Thus, the least distance of the line from the circle is:
\[ \boxed{2\sqrt{5}} \]