The radius of the circle which has normals `xy-2x-y+2= 0` and a tangent `3x+4y-6= 0` is

Options

A. `x^(2)+y^(2)-2x-4y+4=0`
B. `x^(2)+y^(2)-2x-4y+5=0`
C. `x^(2)+y^(2)=5`
D. `(x-3)^(2)+(y-4)^(2)=5`

Explanation

To find the radius of the circle that has the given normals and tangent, we can follow these steps:

### Step 1: Identify the equations of the normals and tangent
The equations provided are:
- Normals: \( xy - 2x - y + 2 = 0 \)
- Tangent: \( 3x + 4y - 6 = 0 \)

### Step 2: Simplify the normal equation
We can rewrite the normal equation:
\[ xy - 2x - y + 2 = 0 \]
Rearranging gives:
\[ xy - 2x - y + 2 = 0 \implies y(x - 1) = 2x - 2 \]
This can be factored as:
\[ (x - 1)(y - 2) = 0 \]
Thus, we have two lines:
1. \( x = 1 \)
2. \( y = 2 \)

### Step 3: Find the center of the circle
The intersection of the lines \( x = 1 \) and \( y = 2 \) gives us the center of the circle:
\[ \text{Center} = (1, 2) \]

### Step 4: Use the distance formula to find the radius
The radius of the circle can be found using the perpendicular distance from the center to the tangent line. The formula for the distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the tangent line \( 3x + 4y - 6 = 0 \), we have:
- \( A = 3 \)
- \( B = 4 \)
- \( C = -6 \)

Substituting the center \( (1, 2) \) into the formula:
\[ d = \frac{|3(1) + 4(2) - 6|}{\sqrt{3^2 + 4^2}} \]

### Step 5: Calculate the numerator
Calculating the numerator:
\[ 3(1) + 4(2) - 6 = 3 + 8 - 6 = 5 \]
Thus, the absolute value is:
\[ |5| = 5 \]

### Step 6: Calculate the denominator
Calculating the denominator:
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 7: Calculate the radius
Now substituting back into the distance formula:
\[ d = \frac{5}{5} = 1 \]
Thus, the radius of the circle is:
\[ \text{Radius} = 1 \]

### Final Answer:
The radius of the circle is \( \boxed{1} \).