The common tangent at the point of contact of the two circles `x^2+y^2-4x-4y=0, x^2+y^2+2x+2y=0` is

Options

A. x+y=0
B. x-y=0
C. 2x-3y=0
D. x-2y=0

Answer

To find the common tangent at the point of contact of the two circles given by the equations \(x^2 + y^2 - 4x - 4y = 0\) and \(x^2 + y^2 + 2x + 2y = 0\), we will follow these steps:

### Step 1: Rewrite the equations of the circles in standard form
The first circle can be rewritten as follows:
\[ x^2 + y^2 - 4x - 4y = 0 \implies (x^2 - 4x) + (y^2 - 4y) = 0 \]
Completing the square:
\[ (x - 2)^2 - 4 + (y - 2)^2 - 4 = 0 \implies (x - 2)^2 + (y - 2)^2 = 8 \]
This represents a circle centered at \((2, 2)\) with radius \(2\sqrt{2}\).

For the second circle:
\[ x^2 + y^2 + 2x + 2y = 0 \implies (x^2 + 2x) + (y^2 + 2y) = 0 \]
Completing the square:
\[ (x + 1)^2 - 1 + (y + 1)^2 - 1 = 0 \implies (x + 1)^2 + (y + 1)^2 = 2 \]
This represents a circle centered at \((-1, -1)\) with radius \(\sqrt{2}\).

### Step 2: Set up the equation for the common tangent
The common tangent condition is given by:
\[ s_2 - s_1 = 0 \]
Where \(s_1\) and \(s_2\) are the equations of the circles. We can express this as:
\[ (x^2 + y^2 + 2x + 2y) - (x^2 + y^2 - 4x - 4y) = 0 \]

### Step 3: Simplify the equation
Simplifying the above equation:
\[ (x^2 + y^2 + 2x + 2y) - (x^2 + y^2 - 4x - 4y) = 0 \]
\[ 2x + 2y + 4x + 4y = 0 \]
\[ 6x + 6y = 0 \]
Dividing through by 6 gives:
\[ x + y = 0 \]

### Conclusion
Thus, the equation of the common tangent at the point of contact of the two circles is:
\[ \boxed{x + y = 0} \]