The tangent to the circle `x^(2)+y^(2)-4x+2y+k=0` at (1,1) is `x-2y+1=0` then k=

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to find the value of \( k \) such that the tangent to the circle at the point (1, 1) is given by the equation \( x - 2y + 1 = 0 \). The equation of the circle is given as \( x^2 + y^2 - 4x + 2y + k = 0 \).

### Step 1: Rewrite the circle equation
The equation of the circle can be rewritten in the standard form:
\[ x^2 + y^2 - 4x + 2y + k = 0 \]
This can be compared with the standard form of a circle:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From this, we identify:
- \( 2g = -4 \) which gives \( g = -2 \)
- \( 2f = 2 \) which gives \( f = 1 \)
- \( c = k \)

### Step 2: Find the center of the circle
The center of the circle \( (h, k) \) can be found using:
\[ h = -g = 2, \quad k = -f = -1 \]
Thus, the center of the circle is \( (2, -1) \).

### Step 3: Calculate the radius of the circle
The radius \( r \) of the circle can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values we have:
\[ r = \sqrt{(-2)^2 + (1)^2 - k} = \sqrt{4 + 1 - k} = \sqrt{5 - k} \]

### Step 4: Find the distance from the center to the point (1, 1)
The distance \( d \) from the center \( (2, -1) \) to the point \( (1, 1) \) is given by the distance formula:
\[ d = \sqrt{(1 - 2)^2 + (1 + 1)^2} = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 5: Set the radius equal to the distance
Since the tangent at the point (1, 1) means that the distance from the center to the point must equal the radius:
\[ \sqrt{5 - k} = \sqrt{5} \]

### Step 6: Square both sides
Squaring both sides gives:
\[ 5 - k = 5 \]

### Step 7: Solve for k
Rearranging the equation yields:
\[ -k = 0 \implies k = 0 \]

### Conclusion
Thus, the value of \( k \) is \( 0 \).