The shortest distance from the line `3x+4y=25` to the circle `x^2+y^2=6x-8y` is equal to :

Options

A. `(7)/(3)`
B. `(9)/(5)`
C. `(11)/(5)`
D. `(7)/(5)`

Answer

To find the shortest distance from the line \(3x + 4y = 25\) to the circle given by the equation \(x^2 + y^2 = 6x - 8y\), we can follow these steps:

### Step 1: Rewrite the equation of the circle
The equation of the circle is given as:
\[ x^2 + y^2 = 6x - 8y \]
We can rearrange this to standard form:
\[ x^2 - 6x + y^2 + 8y = 0 \]
Now, we complete the square for both \(x\) and \(y\).

### Step 2: Complete the square
For \(x^2 - 6x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]
For \(y^2 + 8y\):
\[ y^2 + 8y = (y + 4)^2 - 16 \]
Substituting these back into the equation gives:
\[ (x - 3)^2 - 9 + (y + 4)^2 - 16 = 0 \]
This simplifies to:
\[ (x - 3)^2 + (y + 4)^2 = 25 \]
Thus, the center of the circle is \((3, -4)\) and the radius \(r = 5\).

### Step 3: Find the distance from the center of the circle to the line
The distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(3x + 4y - 25 = 0\), we have \(A = 3\), \(B = 4\), and \(C = -25\). The center of the circle is \((3, -4)\).

Calculating the distance:
\[ d = \frac{|3(3) + 4(-4) - 25|}{\sqrt{3^2 + 4^2}} = \frac{|9 - 16 - 25|}{\sqrt{9 + 16}} = \frac{|-32|}{5} = \frac{32}{5} \]

### Step 4: Calculate the shortest distance from the line to the circle
The shortest distance from the line to the circle is given by:
\[ \text{Shortest Distance} = d - r = \frac{32}{5} - 5 \]
Converting \(5\) to a fraction:
\[ 5 = \frac{25}{5} \]
Thus:
\[ \text{Shortest Distance} = \frac{32}{5} - \frac{25}{5} = \frac{7}{5} \]

### Final Answer
The shortest distance from the line \(3x + 4y = 25\) to the circle is:
\[ \frac{7}{5} \]