The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1, -3). Then its radius is 0

Options

A. `3sqrt2`
B. `2sqrt2`
C. 2
D. 3

Explanation

To find the radius of the circle that touches the line \( x = y \) at the point \( (1, 1) \) and passes through the point \( (1, -3) \), we can follow these steps:

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

### Step 2: Use the point of tangency
Since the circle touches the line \( x = y \) at the point \( (1, 1) \), we can substitute this point into the circle's equation:
\[ (1 - h)^2 + (1 - k)^2 = r^2 \tag{1} \]

### Step 3: Use the point through which the circle passes
The circle also passes through the point \( (1, -3) \). Substituting this point into the circle's equation gives:
\[ (1 - h)^2 + (-3 - k)^2 = r^2 \tag{2} \]

### Step 4: Set up the equations
From equations (1) and (2), we have two equations:
1. \((1 - h)^2 + (1 - k)^2 = r^2\)
2. \((1 - h)^2 + (-3 - k)^2 = r^2\)

Since both equal \( r^2 \), we can set them equal to each other:
\[ (1 - h)^2 + (1 - k)^2 = (1 - h)^2 + (-3 - k)^2 \]

### Step 5: Simplify the equation
Cancelling \((1 - h)^2\) from both sides, we have:
\[ (1 - k)^2 = (-3 - k)^2 \]

### Step 6: Expand both sides
Expanding both sides gives:
\[ 1 - 2k + k^2 = 9 + 6k + k^2 \]

### Step 7: Simplify further
Cancelling \( k^2 \) from both sides:
\[ 1 - 2k = 9 + 6k \]
Rearranging gives:
\[ 1 - 9 = 6k + 2k \]
\[ -8 = 8k \]
\[ k = -1 \]

### Step 8: Substitute \( k \) back to find \( h \)
Now, substitute \( k = -1 \) back into equation (1):
\[ (1 - h)^2 + (1 - (-1))^2 = r^2 \]
This simplifies to:
\[ (1 - h)^2 + 4 = r^2 \]

### Step 9: Substitute \( k \) into equation (2)
Using equation (2):
\[ (1 - h)^2 + (-3 - (-1))^2 = r^2 \]
This simplifies to:
\[ (1 - h)^2 + 4 = r^2 \]

### Step 10: Find the radius
Both equations yield the same result, so we can find \( r \):
\[ r^2 = (1 - h)^2 + 4 \]

### Step 11: Use the condition of tangency
The distance from the center \( (h, k) \) to the line \( x = y \) must equal the radius \( r \). The distance \( d \) from a point \( (h, k) \) to the line \( ax + by + c = 0 \) is given by:
\[ d = \frac{|h - k|}{\sqrt{2}} \]
Setting this equal to \( r \):
\[ \frac{|h - (-1)|}{\sqrt{2}} = r \]

### Step 12: Solve for \( r \)
Substituting \( k = -1 \) into the distance equation and solving will yield the radius \( r \).

### Final Result
After solving, we find that the radius \( r = 2\sqrt{2} \).