Equation of the circle through the origin and making intercepts of 3 and 4 on the positive sides of the axes is

Options

A. `x^(2)+y^(2)+3x+4y=0`
B. `x^(2)+y^(2)-3x-4y=0`
C. `x^(2)+y^(2)+3x-4y=0`
D. `x^(2)+y^(2)-3x+4y=0`

Answer

To find the equation of the circle that passes through the origin and makes intercepts of 3 and 4 on the positive sides of the axes, we can follow these steps:

### Step 1: Understand the general equation of a circle
The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Step 2: Use the condition that the circle passes through the origin
Since the circle passes through the origin \((0, 0)\), we can substitute these coordinates into the general equation:
\[ (0 - h)^2 + (0 - k)^2 = r^2 \]
This simplifies to:
\[ h^2 + k^2 = r^2 \tag{1} \]

### Step 3: Use the intercepts on the axes
The circle makes intercepts of 3 and 4 on the positive sides of the axes. This means it intersects the x-axis at \((3, 0)\) and the y-axis at \((0, 4)\).

### Step 4: Substitute the point \((3, 0)\) into the circle's equation
Substituting \((3, 0)\) into the circle's equation:
\[ (3 - h)^2 + (0 - k)^2 = r^2 \]
This expands to:
\[ (3 - h)^2 + k^2 = r^2 \tag{2} \]

### Step 5: Substitute the point \((0, 4)\) into the circle's equation
Now, substituting \((0, 4)\) into the circle's equation:
\[ (0 - h)^2 + (4 - k)^2 = r^2 \]
This expands to:
\[ h^2 + (4 - k)^2 = r^2 \tag{3} \]

### Step 6: Set equations (1), (2), and (3) equal to each other
From equations (1), (2), and (3), we have:
1. \(h^2 + k^2 = r^2\)
2. \((3 - h)^2 + k^2 = r^2\)
3. \(h^2 + (4 - k)^2 = r^2\)

### Step 7: Equate equations (1) and (2)
Setting (1) equal to (2):
\[ h^2 + k^2 = (3 - h)^2 + k^2 \]
Cancelling \(k^2\) from both sides gives:
\[ h^2 = (3 - h)^2 \]
Expanding the right side:
\[ h^2 = 9 - 6h + h^2 \]
Cancelling \(h^2\) from both sides:
\[ 0 = 9 - 6h \]
Solving for \(h\):
\[ 6h = 9 \implies h = \frac{3}{2} \]

### Step 8: Equate equations (1) and (3)
Now, setting (1) equal to (3):
\[ h^2 + k^2 = h^2 + (4 - k)^2 \]
Cancelling \(h^2\) gives:
\[ k^2 = (4 - k)^2 \]
Expanding the right side:
\[ k^2 = 16 - 8k + k^2 \]
Cancelling \(k^2\) from both sides:
\[ 0 = 16 - 8k \]
Solving for \(k\):
\[ 8k = 16 \implies k = 2 \]

### Step 9: Find the radius \(r\)
Now, substitute \(h\) and \(k\) back into equation (1) to find \(r^2\):
\[ r^2 = h^2 + k^2 = \left(\frac{3}{2}\right)^2 + 2^2 = \frac{9}{4} + 4 = \frac{9}{4} + \frac{16}{4} = \frac{25}{4} \]

### Step 10: Write the equation of the circle
Now we can write the equation of the circle using the values of \(h\), \(k\), and \(r\):
\[ \left(x - \frac{3}{2}\right)^2 + (y - 2)^2 = \frac{25}{4} \]

### Step 11: Expand and simplify the equation
Expanding this gives:
\[ \left(x^2 - 3x + \frac{9}{4}\right) + \left(y^2 - 4y + 4\right) = \frac{25}{4} \]
Combining terms:
\[ x^2 + y^2 - 3x - 4y + \frac{9}{4} + 4 - \frac{25}{4} = 0 \]
This simplifies to:
\[ x^2 + y^2 - 3x - 4y = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 3x - 4y = 0 \]