If the line `a x+b y=2` is a normal to the circle `x^2+y^2-4x-4y=0` and a tangent to the circle `x^2+y^2=1` , then

Options

A. `a=(1)/(2),b=(1)/(2)`
B. `a=(1+sqrt(7))/(2),b=(1-sqrt(7))/(2)`
C. `a=(1)/(4),b=(3)/(4)`
D. `a=1,b=sqrt(3)`

Explanation

To solve the problem step by step, we need to analyze the given conditions regarding the circles and the line.

### Step 1: Identify the center and radius of the first circle
The equation of the first circle is given as:
\[ x^2 + y^2 - 4x - 4y = 0 \]

We can rewrite this in standard form by completing the square:
1. Rearranging the terms:
\[ (x^2 - 4x) + (y^2 - 4y) = 0 \]
2. Completing the square:
\[ (x - 2)^2 - 4 + (y - 2)^2 - 4 = 0 \]
\[ (x - 2)^2 + (y - 2)^2 = 8 \]

From this, we can see that the center of the circle is \( (2, 2) \) and the radius is \( \sqrt{8} = 2\sqrt{2} \).

### Step 2: Use the normal condition
The line given is:
\[ ax + by = 2 \]

Since this line is a normal to the circle, we can substitute the center of the circle into the line equation:
\[ a(2) + b(2) = 2 \]
This simplifies to:
\[ 2a + 2b = 2 \]
Dividing through by 2 gives:
\[ a + b = 1 \] (Equation 1)

### Step 3: Identify the second circle's radius
The second circle is given by:
\[ x^2 + y^2 = 1 \]

This is a circle with center \( (0, 0) \) and radius \( 1 \).

### Step 4: Use the tangent condition
The line \( ax + by = 2 \) is tangent to the second circle. The distance \( d \) from the center of the second circle \( (0, 0) \) to the line \( ax + by = 2 \) is given by the formula:
\[ d = \frac{|ax_1 + by_1 - c|}{\sqrt{a^2 + b^2}} \]
where \( (x_1, y_1) \) is the center of the circle and \( c \) is the constant from the line equation.

Substituting \( (0, 0) \) into the formula gives:
\[ d = \frac{|0 + 0 - 2|}{\sqrt{a^2 + b^2}} = \frac{2}{\sqrt{a^2 + b^2}} \]

Since this distance must equal the radius of the circle (which is \( 1 \)), we have:
\[ \frac{2}{\sqrt{a^2 + b^2}} = 1 \]
Squaring both sides gives:
\[ 4 = a^2 + b^2 \] (Equation 2)

### Step 5: Solve the system of equations
Now we have two equations:
1. \( a + b = 1 \) (Equation 1)
2. \( a^2 + b^2 = 4 \) (Equation 2)

From Equation 1, we can express \( b \) in terms of \( a \):
\[ b = 1 - a \]

Substituting this into Equation 2:
\[ a^2 + (1 - a)^2 = 4 \]
Expanding this gives:
\[ a^2 + (1 - 2a + a^2) = 4 \]
\[ 2a^2 - 2a + 1 = 4 \]
\[ 2a^2 - 2a - 3 = 0 \]
Dividing through by 2:
\[ a^2 - a - \frac{3}{2} = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -1, c = -\frac{3}{2} \):
\[ a = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-\frac{3}{2})}}{2(1)} \]
\[ a = \frac{1 \pm \sqrt{1 + 6}}{2} \]
\[ a = \frac{1 \pm \sqrt{7}}{2} \]

### Step 7: Find \( b \)
Using \( b = 1 - a \):
1. For \( a = \frac{1 + \sqrt{7}}{2} \):
\[ b = 1 - \frac{1 + \sqrt{7}}{2} = \frac{1 - \sqrt{7}}{2} \]

2. For \( a = \frac{1 - \sqrt{7}}{2} \):
\[ b = 1 - \frac{1 - \sqrt{7}}{2} = \frac{1 + \sqrt{7}}{2} \]

Thus, the values of \( a \) and \( b \) are:
- \( a = \frac{1 + \sqrt{7}}{2}, b = \frac{1 - \sqrt{7}}{2} \) or
- \( a = \frac{1 - \sqrt{7}}{2}, b = \frac{1 + \sqrt{7}}{2} \)

### Conclusion
The correct option is:
- \( a = \frac{1 + \sqrt{7}}{2}, b = \frac{1 - \sqrt{7}}{2} \)