The circle `x^2+y^2=4` cuts the circle `x^2+y^2+2x+3y-5=0` in `A` and `B`, Then the equation of the circle on `AB` as diameter is

Options

A. `13(x^(2)+y^(2))-4x-6y-50=0`
B. `9(x^(2)+y^(2))+8x-4y+25=0`
C. `x^(2)+y^(2)-5x+2y+72=0`
D. none of these

Answer

To solve the problem of finding the equation of the circle with diameter AB, where A and B are the points of intersection of the circles given by the equations \(x^2 + y^2 = 4\) and \(x^2 + y^2 + 2x + 3y - 5 = 0\), we will follow these steps:

### Step 1: Write the equations of the circles
The first circle is given by:
\[ S_1: x^2 + y^2 - 4 = 0 \]
The second circle can be rearranged as:
\[ S_2: x^2 + y^2 + 2x + 3y - 5 = 0 \]

### Step 2: Use the family of circles concept
The family of circles passing through the intersection points A and B can be expressed as:
\[ S_1 + \lambda S_2 = 0 \]
Substituting \(S_1\) and \(S_2\):
\[ (x^2 + y^2 - 4) + \lambda (x^2 + y^2 + 2x + 3y - 5) = 0 \]
This simplifies to:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + 2\lambda x + 3\lambda y - (4 + 5\lambda) = 0 \]

### Step 3: Identify the radical axis
The radical axis of the two circles can be found by equating \(S_1\) and \(S_2\):
\[ x^2 + y^2 - 4 = x^2 + y^2 + 2x + 3y - 5 \]
This simplifies to:
\[ 2x + 3y - 1 = 0 \]
Thus, the equation of the radical axis is:
\[ 2x + 3y = 1 \]

### Step 4: Find the center of the circle with diameter AB
The center of the circle that has AB as a diameter lies on the radical axis. The center can be expressed as:
\[ \left(-\frac{g}{1 + \lambda}, -\frac{f}{1 + \lambda}\right) \]
where \(g\) and \(f\) are coefficients from the general circle equation.

### Step 5: Find the values of g and f
From the equation derived in Step 2, we have:
\[ g = \frac{\lambda}{1 + \lambda}, \quad f = \frac{3\lambda}{2(1 + \lambda)} \]

### Step 6: Substitute the center into the radical axis equation
Substituting the center into the equation of the radical axis:
\[ 2\left(-\frac{\lambda}{1 + \lambda}\right) + 3\left(-\frac{3\lambda}{2(1 + \lambda)}\right) = 1 \]
This simplifies to:
\[ -\frac{2\lambda + \frac{9\lambda}{2}}{1 + \lambda} = 1 \]

### Step 7: Solve for \(\lambda\)
Multiply through by \(1 + \lambda\) to eliminate the denominator:
\[ -2\lambda - \frac{9\lambda}{2} = 1 + \lambda \]
Combine like terms and solve for \(\lambda\).

### Step 8: Final equation of the circle
Substituting the value of \(\lambda\) back into the general form of the circle equation derived in Step 2 gives us the final equation of the circle with diameter AB.

### Final Answer
The equation of the circle on AB as diameter is:
\[ 13x^2 + 13y^2 - 4x - 6y - 50 = 0 \]