The equation of circle through origin and cutting intercepts of lengths 2 and 3 from the positive sides of x and y axes is

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. none

Answer

To find the equation of a circle that passes through the origin and cuts intercepts of lengths 2 and 3 from the positive sides of the x and y axes respectively, we can follow these steps:

### Step 1: Write the general equation of the circle
The general equation of a circle can be written as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Since the circle passes through the origin (0, 0), we can substitute \(x = 0\) and \(y = 0\) into the equation:
\[ 0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \]
This simplifies to:
\[ c = 0 \]
Thus, the equation reduces to:
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

### Step 2: Determine the x-intercept
The x-intercept of the circle can be found using the formula:
\[ \text{x-intercept} = \frac{-2g}{\sqrt{g^2 - c}} \]
Since \(c = 0\), the formula becomes:
\[ \text{x-intercept} = \frac{-2g}{|g|} \]
Given that the x-intercept is 2, we set up the equation:
\[ \frac{-2g}{|g|} = 2 \]
This implies:
\[ -2g = 2|g| \]
From this, we can derive two cases:
1. If \(g > 0\), then \(-2g = 2g\) which gives \(g = 0\) (not valid).
2. If \(g < 0\), then \(-2g = -2g\) which gives \(g = -1\).

Thus, we have:
\[ g = -1 \]

### Step 3: Determine the y-intercept
The y-intercept can be found using the formula:
\[ \text{y-intercept} = \frac{-2f}{\sqrt{f^2 - c}} \]
Again, since \(c = 0\), the formula simplifies to:
\[ \text{y-intercept} = \frac{-2f}{|f|} \]
Given that the y-intercept is 3, we set up the equation:
\[ \frac{-2f}{|f|} = 3 \]
This leads to two cases:
1. If \(f > 0\), then \(-2f = 3f\) which gives \(f = 0\) (not valid).
2. If \(f < 0\), then \(-2f = -3f\) which gives \(f = -\frac{2}{3}\).

Thus, we have:
\[ f = -\frac{3}{2} \]

### Step 4: Substitute values of g and f into the equation
Now substituting \(g = -1\) and \(f = -\frac{3}{2}\) into the circle's equation:
\[ x^2 + y^2 + 2(-1)x + 2\left(-\frac{3}{2}\right)y = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2x - 3y = 0 \]

### Final Equation
The equation of the circle is:
\[ x^2 + y^2 - 2x - 3y = 0 \]