The set of values of 'a' for which the point (a-1, a+1) lies outside the circle `x^(2)+y^(2)=8` and inside the circle `x^(2)+y^(2)-12x+12y-62=0`, is

Options

A. `(-oo, -sqrt(3) ) uu(sqrt(3), oo)`
B. `(-3sqrt(2), 3 sqrt(2))`
C. `(-3sqrt(2), -sqrt(3))uu(sqrt(3), 3sqrt(2))`
D. none of these

Explanation

To solve the problem, we need to determine the set of values of 'a' for which the point (a-1, a+1) lies outside the circle defined by the equation \(x^2 + y^2 = 8\) and inside the circle defined by the equation \(x^2 + y^2 - 12x + 12y - 62 = 0\).

### Step 1: Determine the first circle's condition
The first circle has the equation:
\[ x^2 + y^2 = 8 \]
We need to check when the point \((a-1, a+1)\) lies outside this circle. The condition for a point \((x_0, y_0)\) to lie outside a circle \(x^2 + y^2 = r^2\) is:
\[ x_0^2 + y_0^2 > r^2 \]
Substituting \(x_0 = a - 1\) and \(y_0 = a + 1\):
\[ (a - 1)^2 + (a + 1)^2 > 8 \]

### Step 2: Expand the inequality
Expanding the left side:
\[ (a - 1)^2 = a^2 - 2a + 1 \]
\[ (a + 1)^2 = a^2 + 2a + 1 \]
Adding these:
\[ (a^2 - 2a + 1) + (a^2 + 2a + 1) = 2a^2 + 2 \]
So the inequality becomes:
\[ 2a^2 + 2 > 8 \]
Subtracting 8 from both sides:
\[ 2a^2 - 6 > 0 \]
Dividing by 2:
\[ a^2 - 3 > 0 \]
This factors to:
\[ (a - \sqrt{3})(a + \sqrt{3}) > 0 \]

### Step 3: Solve the inequality
The solutions to this inequality are:
\[ a < -\sqrt{3} \quad \text{or} \quad a > \sqrt{3} \]

### Step 4: Determine the second circle's condition
The second circle's equation is:
\[ x^2 + y^2 - 12x + 12y - 62 = 0 \]
Rearranging gives:
\[ x^2 - 12x + y^2 + 12y = 62 \]
Completing the square:
\[ (x - 6)^2 + (y + 6)^2 = 100 \]
This is a circle centered at \((6, -6)\) with a radius of \(10\).

We need to check when the point \((a-1, a+1)\) lies inside this circle. The condition for a point to lie inside a circle is:
\[ (x - h)^2 + (y - k)^2 < r^2 \]
where \((h, k)\) is the center and \(r\) is the radius. Thus, we need:
\[ ((a - 1) - 6)^2 + ((a + 1) + 6)^2 < 100 \]
This simplifies to:
\[ (a - 7)^2 + (a + 7)^2 < 100 \]

### Step 5: Expand the inequality
Expanding:
\[ (a - 7)^2 = a^2 - 14a + 49 \]
\[ (a + 7)^2 = a^2 + 14a + 49 \]
Adding these:
\[ (a^2 - 14a + 49) + (a^2 + 14a + 49) = 2a^2 + 98 \]
So the inequality becomes:
\[ 2a^2 + 98 < 100 \]
Subtracting 98 from both sides:
\[ 2a^2 < 2 \]
Dividing by 2:
\[ a^2 < 1 \]

### Step 6: Solve the inequality
The solutions to this inequality are:
\[ -\sqrt{1} < a < \sqrt{1} \quad \Rightarrow \quad -1 < a < 1 \]

### Step 7: Combine the results
Now we combine the results from both inequalities:
1. From the first circle: \(a < -\sqrt{3}\) or \(a > \sqrt{3}\)
2. From the second circle: \(-1 < a < 1\)

The intersection of these conditions gives us the final set of values for \(a\):
\[ a \in (-1, 1) \cap (-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty) \]

Since \(-\sqrt{3} \approx -1.732\) and \(\sqrt{3} \approx 1.732\), the final intervals are:
\[ a \in (-1, -\sqrt{3}) \cup (\sqrt{3}, 1) \]

### Final Answer
The set of values of \(a\) is:
\[ (-1, -\sqrt{3}) \cup (\sqrt{3}, 1) \]