Find the equation of a circle which is co-axial with the circles `x^(2)+y^(2)+4x+2y+1=0andx^(2)+y^(2)-x+3y-(3)/(2)=0` and having its centre on the radical axis of these circles.

Answer

To find the equation of a circle that is coaxial with the given circles and has its center on the radical axis, we will follow these steps:

### Step 1: Identify the given circles
The equations of the circles are:
1. \( C_1: x^2 + y^2 + 4x + 2y + 1 = 0 \)
2. \( C_2: x^2 + y^2 - x + 3y - \frac{3}{2} = 0 \)

### Step 2: Write the equations in standard form
We can rewrite the equations of the circles in the standard form \( (x - h)^2 + (y - k)^2 = r^2 \).

For \( C_1 \):
\[ x^2 + 4x + y^2 + 2y + 1 = 0 \]
Completing the square:
\[ (x^2 + 4x + 4) + (y^2 + 2y + 1) = 4 \]
\[ (x + 2)^2 + (y + 1)^2 = 4 \]
This circle has center \( (-2, -1) \) and radius \( 2 \).

For \( C_2 \):
\[ x^2 - x + y^2 + 3y - \frac{3}{2} = 0 \]
Completing the square:
\[ (x^2 - x + \frac{1}{4}) + (y^2 + 3y + \frac{9}{4}) = \frac{1}{4} + \frac{9}{4} + \frac{3}{2} \]
\[ (x - \frac{1}{2})^2 + (y + \frac{3}{2})^2 = 4 \]
This circle has center \( \left(\frac{1}{2}, -\frac{3}{2}\right) \) and radius \( 2 \).

### Step 3: Find the radical axis
The radical axis of two circles is given by the equation \( S_1 - S_2 = 0 \), where \( S_1 \) and \( S_2 \) are the left-hand sides of the equations of the circles.

Calculating \( S_1 - S_2 \):
\[ S_1 = x^2 + y^2 + 4x + 2y + 1 \]
\[ S_2 = x^2 + y^2 - x + 3y - \frac{3}{2} \]
Subtracting:
\[ S_1 - S_2 = (4x + 2y + 1) - (-x + 3y - \frac{3}{2}) = 5x - y + \frac{5}{2} = 0 \]
Rearranging gives:
\[ 10x - 2y + 5 = 0 \quad \text{(Equation of the radical axis)} \]

### Step 4: Form the equation of the coaxial circle
The general form of the equation of a circle coaxial with the given circles is:
\[ S_1 + \lambda S_2 = 0 \]
Substituting \( S_1 \) and \( S_2 \):
\[ x^2 + y^2 + 4x + 2y + 1 + \lambda (x^2 + y^2 - x + 3y - \frac{3}{2}) = 0 \]
Combining like terms:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + (4 - \lambda)x + (2 + 3\lambda)y + (1 - \frac{3\lambda}{2}) = 0 \]

### Step 5: Find the center of the coaxial circle
The center of the circle can be found from the coefficients:
\[ g = \frac{-(4 - \lambda)}{2(1 + \lambda)}, \quad f = \frac{-(2 + 3\lambda)}{2(1 + \lambda)} \]
Setting the center on the radical axis:
\[ 10g - 2f + 5 = 0 \]
Substituting \( g \) and \( f \):
\[ 10\left(\frac{-(4 - \lambda)}{2(1 + \lambda)}\right) - 2\left(\frac{-(2 + 3\lambda)}{2(1 + \lambda)}\right) + 5 = 0 \]
Simplifying:
\[ -5(4 - \lambda) + (2 + 3\lambda) + 5(1 + \lambda) = 0 \]
\[ -20 + 5\lambda + 2 + 3\lambda + 5 + 5\lambda = 0 \]
\[ 13\lambda - 13 = 0 \implies \lambda = 1 \]

### Step 6: Substitute \( \lambda \) back to find the circle
Substituting \( \lambda = 1 \) into the coaxial circle equation:
\[ S_1 + S_2 = 0 \]
\[ x^2 + y^2 + 4x + 2y + 1 + (x^2 + y^2 - x + 3y - \frac{3}{2}) = 0 \]
Combining gives:
\[ 2x^2 + 2y^2 + 3x + 5y - \frac{1}{2} = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 4x^2 + 4y^2 + 6x + 10y - 1 = 0 \]

### Final Equation of the Circle
Thus, the equation of the required circle is:
\[ 4x^2 + 4y^2 + 6x + 10y - 1 = 0 \]