Find the coordinates of the limiting points of the system of circles determined by the two cricles
`x^(2)+y^(2)+5x+y+4=0andx^(2)+y^(2)+10x-4y-1=0`

Answer

To find the coordinates of the limiting points of the system of circles determined by the two given circles, we will follow these steps:

### Step 1: Write the equations of the circles
The equations of the circles are:
1. \( x^2 + y^2 + 5x + y + 4 = 0 \) (let's call this Circle 1)
2. \( x^2 + y^2 + 10x - 4y - 1 = 0 \) (let's call this Circle 2)

### Step 2: Find the common chord
To find the common chord, we use the formula:
\[ S - S_1 - S_2 = 0 \]
where \( S \) is the general equation of the circle, and \( S_1 \) and \( S_2 \) are the equations of the two circles.

### Step 3: Substitute the equations into the common chord formula
Let:
- \( S_1 = x^2 + y^2 + 5x + y + 4 \)
- \( S_2 = x^2 + y^2 + 10x - 4y - 1 \)

Now, we substitute these into the common chord formula:
\[ S - (S_1 + S_2) = 0 \]
This gives us:
\[ S - (x^2 + y^2 + 5x + y + 4 + x^2 + y^2 + 10x - 4y - 1) = 0 \]
Simplifying this, we have:
\[ S - (2x^2 + 2y^2 + 15x - 3y + 3) = 0 \]

### Step 4: Simplify to find the common chord equation
This simplifies to:
\[ 0 = 2x^2 + 2y^2 + 15x - 3y + 3 \]
Now, we can divide the entire equation by 2:
\[ 0 = x^2 + y^2 + \frac{15}{2}x - \frac{3}{2}y + \frac{3}{2} \]

### Step 5: Set up the limiting points equation
The limiting points can be found by equating the general circle equation \( S \) to a linear combination of the two circle equations:
\[ S = S_1 + \lambda (S_2 - S_1) \]
This leads to:
\[ x^2 + y^2 + 5x + y + 4 + \lambda (5x - 5y + 5) = 0 \]

### Step 6: Collect terms
Collecting terms gives:
\[ x^2 + y^2 + (5 + 5\lambda)x + (1 - 5\lambda)y + (4 + 5\lambda) = 0 \]

### Step 7: Find the radius condition
For the circles to have limiting points, the radius must be zero:
\[ \left(\frac{5 + 5\lambda}{2}\right)^2 + \left(\frac{1 - 5\lambda}{2}\right)^2 - (4 + 5\lambda) = 0 \]

### Step 8: Solve for \(\lambda\)
Expanding and simplifying:
\[ \frac{(5 + 5\lambda)^2 + (1 - 5\lambda)^2}{4} - (4 + 5\lambda) = 0 \]
This leads to a quadratic in \(\lambda\):
\[ 2\lambda^2 + 12\lambda + 10 = 0 \]

### Step 9: Use the quadratic formula
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-12 \pm \sqrt{144 - 80}}{4} = \frac{-12 \pm 8}{4} \]
This gives:
\[ \lambda_1 = -1, \quad \lambda_2 = -5 \]

### Step 10: Find the coordinates of the limiting points
Substituting \(\lambda_1\) and \(\lambda_2\) back into the center coordinates:
1. For \(\lambda_1 = -1\):
\[ C_1 = \left(5 - 1 - 2, 1 + 1 - 2\right) = (-2, -1) \]
2. For \(\lambda_2 = -5\):
\[ C_2 = \left(5 - 5 - 2, 1 + 5 - 2\right) = (0, -3) \]

### Final Answer
The coordinates of the limiting points are:
1. \( (-2, -1) \)
2. \( (0, -3) \)