Equation of a circle which passes through `(3,6)` and touches the axes is

Options

A. `x^(2)+y^(2)+6x+6y+3=0`
B. `x^(2)+y^(2)-6x-6y-9=0`
C. `x^(2)+y^(2)-6x-6y+9=0`
D. `"none of these"`

Explanation

To find the equation of a circle that passes through the point (3, 6) and touches both the x-axis and y-axis, we can follow these steps:

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

### Step 2: Write the general 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, since the center is (r, r), the equation becomes:
\[ (x - r)^2 + (y - r)^2 = r^2 \]

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

### Step 4: Expand and simplify the equation
Now we expand the left-hand side:
\[ (3 - r)^2 = 9 - 6r + r^2 \]
\[ (6 - r)^2 = 36 - 12r + r^2 \]
Adding these together gives:
\[ 9 - 6r + r^2 + 36 - 12r + r^2 = r^2 \]
Combining like terms:
\[ 2r^2 - 18r + 45 = r^2 \]
Subtract \(r^2\) from both sides:
\[ r^2 - 18r + 45 = 0 \]

### Step 5: Solve the quadratic equation
Now we can solve the quadratic equation \(r^2 - 18r + 45 = 0\) using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -18\), and \(c = 45\):
\[ r = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1} \]
\[ r = \frac{18 \pm \sqrt{324 - 180}}{2} \]
\[ r = \frac{18 \pm \sqrt{144}}{2} \]
\[ r = \frac{18 \pm 12}{2} \]
Calculating the two possible values:
1. \(r = \frac{30}{2} = 15\)
2. \(r = \frac{6}{2} = 3\)

### Step 6: Write the equations of the circles
Now we have two possible values for \(r\):
1. For \(r = 15\):
\[ (x - 15)^2 + (y - 15)^2 = 15^2 \]
Expanding this gives:
\[ (x - 15)^2 + (y - 15)^2 = 225 \]

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

### Step 7: Final equations
Thus, the equations of the circles are:
1. \( (x - 15)^2 + (y - 15)^2 = 225 \)
2. \( (x - 3)^2 + (y - 3)^2 = 9 \)