The equation of circles passing through `(3, -6)` touching both the axes is

Options

A. `x^(2)+y^(2)-6x+6y+9=0`
B. `x^(2)+y^(2)+6x-6y+9=0`
C. `x^(2)+y^(2)+30x - 30y+225=0`
D. `x^(2)+y^(2)+30x +30y+225=0`

Answer

To find the equation of circles passing through the point (3, -6) and touching both the axes, we can follow these steps:

### Step 1: Understand the properties of the circle
Since the circle touches both the x-axis and y-axis, the center of the circle must be at a point (r, -r) where r is the radius of the circle. This is because the distance from the center to the axes must equal the radius.

### Step 2: Set up the equation of the circle
The general equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
In our case, the center is (r, -r), so the equation becomes:
\[ (x - r)^2 + (y + r)^2 = r^2 \]

### Step 3: Substitute the point (3, -6)
Since the circle passes through the point (3, -6), we can substitute these coordinates into the equation:
\[ (3 - r)^2 + (-6 + r)^2 = r^2 \]

### Step 4: Expand the equation
Now, we expand both sides:
1. Expanding \((3 - r)^2\):
\[ (3 - r)^2 = 9 - 6r + r^2 \]
2. Expanding \((-6 + r)^2\):
\[ (-6 + r)^2 = 36 - 12r + r^2 \]

Putting it all together, we have:
\[ 9 - 6r + r^2 + 36 - 12r + r^2 = r^2 \]

### Step 5: Combine like terms
Combine the terms on the left side:
\[ 2r^2 - 18r + 45 = r^2 \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ 2r^2 - r^2 - 18r + 45 = 0 \implies r^2 - 18r + 45 = 0 \]

### Step 7: Solve the quadratic equation
Now we can solve the quadratic equation using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -18\), and \(c = 45\):
\[ r = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1} \]
Calculating the discriminant:
\[ = \frac{18 \pm \sqrt{324 - 180}}{2} = \frac{18 \pm \sqrt{144}}{2} = \frac{18 \pm 12}{2} \]
This gives us two possible values for r:
1. \(r = \frac{30}{2} = 15\)
2. \(r = \frac{6}{2} = 3\)

### Step 8: Write the equations of the circles
Now we substitute these values of r back into the circle equation:
1. For \(r = 15\):
\[ (x - 15)^2 + (y + 15)^2 = 225 \]
Expanding this gives:
\[ x^2 - 30x + 225 + y^2 + 30y + 225 = 225 \implies x^2 + y^2 - 30x + 30y + 225 = 0 \]

2. For \(r = 3\):
\[ (x - 3)^2 + (y + 3)^2 = 9 \]
Expanding this gives:
\[ x^2 - 6x + 9 + y^2 + 6y + 9 = 9 \implies x^2 + y^2 - 6x + 6y + 9 = 0 \]

### Final Answer
The equations of the circles passing through (3, -6) and touching both axes are:
1. \(x^2 + y^2 - 30x + 30y + 225 = 0\)
2. \(x^2 + y^2 - 6x + 6y + 9 = 0\)