What is the shortest distance between the circle `x^(2)+y^(2)-2x-2y=0` and the line x=4?

Options

A. 1.23
B. 1.59
C. 2.56
D. 3.25

Answer

To find the shortest distance between the circle given by the equation \(x^2 + y^2 - 2x - 2y = 0\) and the line \(x = 4\), we will follow these steps:

### Step 1: Rewrite the Circle's Equation in Standard Form
The given equation of the circle is:
\[ x^2 + y^2 - 2x - 2y = 0 \]
We can rearrange this by grouping the \(x\) and \(y\) terms:
\[ (x^2 - 2x) + (y^2 - 2y) = 0 \]
Next, we complete the square for both \(x\) and \(y\).

For \(x^2 - 2x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
For \(y^2 - 2y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]
Substituting back, we have:
\[ ((x - 1)^2 - 1) + ((y - 1)^2 - 1) = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y - 1)^2 - 2 = 0 \]
Thus, we can rewrite it as:
\[ (x - 1)^2 + (y - 1)^2 = 2 \]
This is the standard form of the circle, where the center is \((1, 1)\) and the radius \(r\) is \(\sqrt{2}\).

### Step 2: Determine the Distance from the Circle's Center to the Line
The center of the circle is at the point \((1, 1)\). The line \(x = 4\) is a vertical line. To find the distance from the center of the circle to this line, we calculate the horizontal distance between the \(x\)-coordinates:
\[ \text{Distance} = |4 - 1| = 3 \]

### Step 3: Calculate the Shortest Distance from the Circle to the Line
The shortest distance from the circle to the line is the distance from the center to the line minus the radius of the circle:
\[ \text{Shortest Distance} = \text{Distance from center to line} - \text{Radius} \]
Substituting the values:
\[ \text{Shortest Distance} = 3 - \sqrt{2} \]
Calculating \(\sqrt{2} \approx 1.414\):
\[ \text{Shortest Distance} \approx 3 - 1.414 \approx 1.586 \]

### Final Answer
Thus, the shortest distance between the circle and the line \(x = 4\) is approximately \(1.586\).