The line `2x-y+1=0` touches a circle at the point (2, 5) and the centre of the circle lies on `x-2y=4`. The diameter (in units) of the circle is

Options

A. `6sqrt5`
B. `10sqrt3`
C. `4sqrt5`
D. `10sqrt2`

Answer

To find the diameter of the circle that touches the line \(2x - y + 1 = 0\) at the point \((2, 5)\) and has its center on the line \(x - 2y = 4\), we can follow these steps:

### Step 1: Identify the Tangent Line and Point of Tangency
The given tangent line is:
\[ 2x - y + 1 = 0 \]
This line touches the circle at the point:
\[ (2, 5) \]

### Step 2: Find the Slope of the Tangent Line
Rearranging the equation of the tangent line into slope-intercept form \(y = mx + c\):
\[ y = 2x + 1 \]
The slope \(m_1\) of the tangent line is:
\[ m_1 = 2 \]

### Step 3: Determine the Slope of the Normal Line
The slope of the normal line (which passes through the center of the circle) is the negative reciprocal of the slope of the tangent line:
\[ m_2 = -\frac{1}{m_1} = -\frac{1}{2} \]

### Step 4: Use the Point-Slope Form to Find the Center
Using the point \((2, 5)\) and the slope of the normal line, we can write the equation of the normal line:
\[ y - 5 = -\frac{1}{2}(x - 2) \]
This simplifies to:
\[ y - 5 = -\frac{1}{2}x + 1 \implies y = -\frac{1}{2}x + 6 \]

### Step 5: Find the Intersection with the Line \(x - 2y = 4\)
The center of the circle lies on the line:
\[ x - 2y = 4 \]
We can substitute \(y = -\frac{1}{2}x + 6\) into this equation:
\[ x - 2\left(-\frac{1}{2}x + 6\right) = 4 \]
This simplifies to:
\[ x + x - 12 = 4 \implies 2x - 12 = 4 \implies 2x = 16 \implies x = 8 \]
Now substituting \(x = 8\) back to find \(y\):
\[ y = -\frac{1}{2}(8) + 6 = -4 + 6 = 2 \]
Thus, the center of the circle is:
\[ (8, 2) \]

### Step 6: Calculate the Radius of the Circle
Using the distance formula to find the radius \(r\) between the center \((8, 2)\) and the point of tangency \((2, 5)\):
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(8 - 2)^2 + (2 - 5)^2} \]
Calculating the differences:
\[ = \sqrt{(6)^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \]

### Step 7: Calculate the Diameter of the Circle
The diameter \(D\) of the circle is twice the radius:
\[ D = 2r = 2(3\sqrt{5}) = 6\sqrt{5} \]

Thus, the diameter of the circle is:
\[ \boxed{6\sqrt{5}} \]