If the circle `x^(2)+y^(2)-2x-6y = r^(2)-10` is tangent to the line 12y = 60 , the value of r is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to find the value of \( r \) such that the circle defined by the equation \( x^2 + y^2 - 2x - 6y = r^2 - 10 \) is tangent to the line \( 12y = 60 \).

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 - 2x - 6y = r^2 - 10 \]
We can rearrange this to:
\[ x^2 - 2x + y^2 - 6y = r^2 - 10 \]

### Step 2: Complete the square for \( x \) and \( y \)
To convert the left side into a standard circle equation, we complete the square for both \( x \) and \( y \).

1. For \( x^2 - 2x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

2. For \( y^2 - 6y \):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Substituting these back into the equation gives:
\[ (x - 1)^2 - 1 + (y - 3)^2 - 9 = r^2 - 10 \]
This simplifies to:
\[ (x - 1)^2 + (y - 3)^2 - 10 = r^2 - 10 \]
Thus, we have:
\[ (x - 1)^2 + (y - 3)^2 = r^2 \]

### Step 3: Identify the center and radius of the circle
From the equation \( (x - 1)^2 + (y - 3)^2 = r^2 \), we can see that:
- The center of the circle is \( (1, 3) \).
- The radius of the circle is \( r \).

### Step 4: Determine the equation of the line
The line given is \( 12y = 60 \), which simplifies to:
\[ y = 5 \]
This is a horizontal line at \( y = 5 \).

### Step 5: Calculate the distance from the center to the line
The distance \( d \) from the center of the circle \( (1, 3) \) to the line \( y = 5 \) is given by:
\[ d = |y_1 - y_2| = |3 - 5| = 2 \]

### Step 6: Set the distance equal to the radius
Since the circle is tangent to the line, the distance from the center to the line is equal to the radius \( r \):
\[ r = 2 \]

### Conclusion
Thus, the value of \( r \) is:
\[ \boxed{2} \]