Find the values of a for which the point `(a,a) , a gt 0,` lies outside the circle `x ^(2) + y ^(2) - 2x + 6y -6=0`

Answer

To find the values of \( a \) for which the point \( (a, a) \) lies outside the circle defined by the equation \( x^2 + y^2 - 2x + 6y - 6 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the circle equation in standard form. The given equation is:
\[ x^2 + y^2 - 2x + 6y - 6 = 0 \]
We can rearrange this to:
\[ x^2 - 2x + y^2 + 6y = 6 \]
Next, we complete the square for both \( x \) and \( y \).

### Step 2: Complete the Square
For \( x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
For \( y \):
\[ y^2 + 6y = (y + 3)^2 - 9 \]
Substituting these back into the equation gives:
\[ (x - 1)^2 - 1 + (y + 3)^2 - 9 = 6 \]
Simplifying this, we get:
\[ (x - 1)^2 + (y + 3)^2 - 10 = 6 \]
Thus, we have:
\[ (x - 1)^2 + (y + 3)^2 = 16 \]
This represents a circle with center \( (1, -3) \) and radius \( 4 \).

### Step 3: Substitute the Point into the Circle Equation
Now, we substitute the point \( (a, a) \) into the circle equation. The point lies outside the circle if the value of \( S_1 \) (the left side of the circle equation evaluated at the point) is greater than \( 0 \):
\[ S_1 = (a - 1)^2 + (a + 3)^2 - 16 > 0 \]
Calculating \( S_1 \):
\[ S_1 = (a - 1)^2 + (a + 3)^2 - 16 \]
Expanding the squares:
\[ = (a^2 - 2a + 1) + (a^2 + 6a + 9) - 16 \]
Combining like terms:
\[ = 2a^2 + 4a - 6 > 0 \]

### Step 4: Factor the Quadratic
Now we need to factor the quadratic inequality:
\[ 2a^2 + 4a - 6 > 0 \]
Dividing the entire inequality by \( 2 \):
\[ a^2 + 2a - 3 > 0 \]
Factoring gives:
\[ (a + 3)(a - 1) > 0 \]

### Step 5: Analyze the Inequality
To find the intervals where this inequality holds, we can use a number line:
- The roots of the equation \( (a + 3)(a - 1) = 0 \) are \( a = -3 \) and \( a = 1 \).
- We test intervals: \( (-\infty, -3) \), \( (-3, 1) \), and \( (1, \infty) \).

1. For \( a < -3 \): both factors are negative, product is positive.
2. For \( -3 < a < 1 \): one factor is negative, the other is positive, product is negative.
3. For \( a > 1 \): both factors are positive, product is positive.

Thus, the solution to the inequality is:
\[ a < -3 \quad \text{or} \quad a > 1 \]

### Step 6: Apply the Condition \( a > 0 \)
Since we are looking for values of \( a \) such that \( a > 0 \), we only consider the interval \( (1, \infty) \).

### Final Answer
The values of \( a \) for which the point \( (a, a) \) lies outside the circle are:
\[ a \in (1, \infty) \]