If the origin be one limiting point of system of co-axial circles of which `x^(2)+y^(2)+3x+4y+25=0` is a member ,find the other limiting point.

Answer

To find the other limiting point of the system of co-axial circles given that the origin is one limiting point, we start with the equation of the circle:

**Given Equation:**
\[ x^2 + y^2 + 3x + 4y + 25 = 0 \]

### Step 1: Identify the coefficients
The general form of the equation of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation, we can identify:
- \( 2g = 3 \) → \( g = \frac{3}{2} \)
- \( 2f = 4 \) → \( f = 2 \)
- \( c = 25 \)

### Step 2: Find the center of the circle
The center of the circle is given by the coordinates \((-g, -f)\):
- Center = \((-g, -f) = \left(-\frac{3}{2}, -2\right)\)

### Step 3: Use the formula for limiting points
The limiting points of a system of co-axial circles can be found using the formulas:
- \( x = -\frac{gc}{g^2 + f^2} \)
- \( y = -\frac{fc}{g^2 + f^2} \)

### Step 4: Calculate \( g^2 + f^2 \)
First, we need to calculate \( g^2 + f^2 \):
- \( g^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \)
- \( f^2 = 2^2 = 4 \)
- Therefore, \( g^2 + f^2 = \frac{9}{4} + 4 = \frac{9}{4} + \frac{16}{4} = \frac{25}{4} \)

### Step 5: Calculate the x-coordinate of the limiting point
Now substituting \( g \), \( c \), and \( g^2 + f^2 \) into the x-coordinate formula:
\[ x = -\frac{gc}{g^2 + f^2} = -\frac{\left(\frac{3}{2}\right)(25)}{\frac{25}{4}} = -\frac{\frac{75}{2}}{\frac{25}{4}} = -\frac{75}{2} \cdot \frac{4}{25} = -\frac{75 \cdot 4}{2 \cdot 25} = -\frac{300}{50} = -6 \]

### Step 6: Calculate the y-coordinate of the limiting point
Now substituting \( f \), \( c \), and \( g^2 + f^2 \) into the y-coordinate formula:
\[ y = -\frac{fc}{g^2 + f^2} = -\frac{(2)(25)}{\frac{25}{4}} = -\frac{50}{\frac{25}{4}} = -50 \cdot \frac{4}{25} = -\frac{200}{25} = -8 \]

### Final Result
Thus, the other limiting point is:
\[ \text{Other Limiting Point} = (x, y) = (-6, -8) \]

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