Find the equation of the circle which passes through the origin and cuts off intercepts -2 and 3 from the coordinate axes .

Options

A. `x^(2) + y^(2) + 2x + 3y = 0
B. ` x^(2) + y^(2) + 2x - 3y = 0 `
C. ` x^(2) + y^(2) - 2x + 3y = 0 `
D. ` x^(2) + y^(2) - 2x - 3y = 0 `

Answer

To find the equation of the circle that passes through the origin and cuts off intercepts of -2 and 3 from the coordinate axes, we can follow these steps:

### Step 1: Write the general equation of the circle
The general equation of a circle can be written as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

### Step 2: Use the fact that the circle passes through the origin
Since the circle passes through the origin (0, 0), we substitute \( x = 0 \) and \( y = 0 \) into the equation:
\[ 0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \]
This simplifies to:
\[ c = 0 \]

### Step 3: Substitute the intercepts into the equation
The circle cuts off intercepts of -2 and 3 from the coordinate axes. This means:
- The x-intercept is -2 (when \( y = 0 \)), so we set \( x = -2 \) and \( y = 0 \).
- The y-intercept is 3 (when \( x = 0 \)), so we set \( x = 0 \) and \( y = 3 \).

#### For the x-intercept (-2, 0):
Substituting \( x = -2 \) and \( y = 0 \) into the circle equation:
\[ (-2)^2 + 0^2 + 2g(-2) + 2f(0) + 0 = 0 \]
This simplifies to:
\[ 4 - 4g = 0 \]
Thus, we have:
\[ 4g = 4 \]
\[ g = 1 \]

#### For the y-intercept (0, 3):
Now substituting \( x = 0 \) and \( y = 3 \) into the circle equation:
\[ 0^2 + 3^2 + 2g(0) + 2f(3) + 0 = 0 \]
This simplifies to:
\[ 9 + 6f = 0 \]
Thus, we have:
\[ 6f = -9 \]
\[ f = -\frac{3}{2} \]

### Step 4: Substitute the values of g, f, and c back into the equation
Now we substitute \( g = 1 \), \( f = -\frac{3}{2} \), and \( c = 0 \) back into the general equation:
\[ x^2 + y^2 + 2(1)x + 2\left(-\frac{3}{2}\right)y + 0 = 0 \]
This simplifies to:
\[ x^2 + y^2 + 2x - 3y = 0 \]

### Final Answer
Thus, the required equation of the circle is:
\[ x^2 + y^2 + 2x - 3y = 0 \]

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