Find the coordinates of the middle point of the chord which the circle `x^(2)+y^(2)+4x-2y-3=0` cuts-off the line x-y+2=0.

Answer

To find the coordinates of the midpoint of the chord that the circle \(x^2 + y^2 + 4x - 2y - 3 = 0\) cuts off from the line \(x - y + 2 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
The given circle equation is:
\[ x^2 + y^2 + 4x - 2y - 3 = 0 \]
We can rewrite this in the standard form of a circle \( (x - h)^2 + (y - k)^2 = r^2 \).

### Step 2: Identify the Center and Radius of the Circle
To find the center and radius, we can complete the square for the \(x\) and \(y\) terms.

1. Rearranging:
\[ (x^2 + 4x) + (y^2 - 2y) = 3 \]

2. Completing the square:
- For \(x^2 + 4x\):
\[ x^2 + 4x = (x + 2)^2 - 4 \]
- For \(y^2 - 2y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

3. Substituting back:
\[ (x + 2)^2 - 4 + (y - 1)^2 - 1 = 3 \]
\[ (x + 2)^2 + (y - 1)^2 = 8 \]

Thus, the center of the circle is \((-2, 1)\) and the radius is \(\sqrt{8} = 2\sqrt{2}\).

### Step 3: Find the Equation of the Line
The line given is:
\[ x - y + 2 = 0 \]
Rearranging gives:
\[ y = x + 2 \]

### Step 4: Find the Slope of the Line
The slope of this line is \(m_1 = 1\).

### Step 5: Find the Perpendicular Line from the Center to the Given Line
The slope of the line perpendicular to the given line is:
\[ m_2 = -\frac{1}{m_1} = -1 \]

Using the point-slope form of the line equation:
\[ y - 1 = -1(x + 2) \]
This simplifies to:
\[ y = -x - 1 \]

### Step 6: Find the Intersection Point (Midpoint of the Chord)
To find the intersection of the two lines \(y = x + 2\) and \(y = -x - 1\), we set them equal to each other:
\[ x + 2 = -x - 1 \]
Solving for \(x\):
\[ 2x = -3 \implies x = -\frac{3}{2} \]

Substituting \(x\) back into one of the line equations to find \(y\):
\[ y = -\left(-\frac{3}{2}\right) - 1 = \frac{3}{2} - 1 = \frac{1}{2} \]

### Final Coordinates of the Midpoint
Thus, the coordinates of the midpoint \(M\) of the chord are:
\[ M\left(-\frac{3}{2}, \frac{1}{2}\right) \]