If `y=f(x)=ax+b` is a tangent to circle `x^2+y^2+2x+2y-2=0` then the value of `(a+b)^2 +2(a-b)(a+b+1)` is equal to

Options

A. 0
B. 1
C. -3
D. none of these

Answer

To solve the problem, we need to determine the value of \((a+b)^2 + 2(a-b)(a+b+1)\) given that the line \(y = ax + b\) is a tangent to the circle defined by the equation \(x^2 + y^2 + 2x + 2y - 2 = 0\).

### Step 1: Rewrite the Circle Equation
First, we rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 + 2x + 2y - 2 = 0 \]
We can complete the square for \(x\) and \(y\):
\[ (x^2 + 2x) + (y^2 + 2y) = 2 \]
Completing the square:
\[ (x+1)^2 - 1 + (y+1)^2 - 1 = 2 \]
\[ (x+1)^2 + (y+1)^2 = 4 \]
This shows that the circle has center \((-1, -1)\) and radius \(r = 2\).

### Step 2: Find the Perpendicular Distance from the Center to the Line
The line \(y = ax + b\) can be rewritten in the standard form:
\[ ax - y + b = 0 \]
The formula for the perpendicular distance \(d\) from a point \((h, k)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = a\), \(B = -1\), \(C = b\), and the center of the circle is \((-1, -1)\). Substituting these values, we get:
\[ d = \frac{|a(-1) - 1(-1) + b|}{\sqrt{a^2 + (-1)^2}} = \frac{| -a + 1 + b|}{\sqrt{a^2 + 1}} \]

### Step 3: Set the Distance Equal to the Radius
Since the line is a tangent to the circle, the distance \(d\) must equal the radius \(r = 2\):
\[ \frac{| -a + 1 + b|}{\sqrt{a^2 + 1}} = 2 \]
Squaring both sides gives:
\[ (-a + 1 + b)^2 = 4(a^2 + 1) \]

### Step 4: Expand and Rearrange
Expanding both sides:
\[ (-a + 1 + b)^2 = a^2 - 2a + 2b - 2ab + b^2 + 1 \]
And the right side:
\[ 4(a^2 + 1) = 4a^2 + 4 \]
Setting these equal:
\[ (-a + 1 + b)^2 = 4a^2 + 4 \]
This leads us to:
\[ (-a + 1 + b)^2 - 4a^2 - 4 = 0 \]

### Step 5: Solve for \(a\) and \(b\)
This equation can be simplified and rearranged to find a relationship between \(a\) and \(b\). After some algebra, we can derive a quadratic equation in terms of \(a\) and \(b\).

### Step 6: Calculate \((a+b)^2 + 2(a-b)(a+b+1)\)
Now we need to compute the expression:
\[ (a+b)^2 + 2(a-b)(a+b+1) \]
Substituting the values obtained from the previous steps into this expression will yield the final result.

### Final Result
After performing the necessary algebraic manipulations and substitutions, we find that the value of \((a+b)^2 + 2(a-b)(a+b+1)\) simplifies to \(-3\).

Thus, the final answer is:
\[ \boxed{-3} \]