The area of the circle `x^(2)-2x+y^(2)-10y+k=0` is `25pi` sq. units. The value of k is equal to

Options

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

Answer

To find the value of \( k \) in the equation of the circle \( x^2 - 2x + y^2 - 10y + k = 0 \) given that the area of the circle is \( 25\pi \) square units, we can follow these steps:

### Step 1: Rewrite the equation in standard form
The given equation is:
\[ x^2 - 2x + y^2 - 10y + k = 0 \]
We need to complete the square for both \( x \) and \( y \).

### Step 2: Complete the square for \( x \)
For \( x^2 - 2x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

### Step 3: Complete the square for \( y \)
For \( y^2 - 10y \):
\[ y^2 - 10y = (y - 5)^2 - 25 \]

### Step 4: Substitute back into the equation
Now substituting back into the equation:
\[ (x - 1)^2 - 1 + (y - 5)^2 - 25 + k = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y - 5)^2 + k - 26 = 0 \]
Rearranging gives:
\[ (x - 1)^2 + (y - 5)^2 = 26 - k \]

### Step 5: Relate to the area of the circle
The area \( A \) of a circle is given by:
\[ A = \pi r^2 \]
Given that the area is \( 25\pi \), we can set up the equation:
\[ \pi r^2 = 25\pi \]
Dividing both sides by \( \pi \):
\[ r^2 = 25 \]

### Step 6: Set the radius squared equal to the equation
From our earlier rearrangement, we have:
\[ r^2 = 26 - k \]
Setting this equal to \( 25 \):
\[ 26 - k = 25 \]

### Step 7: Solve for \( k \)
Now, solving for \( k \):
\[ k = 26 - 25 \]
\[ k = 1 \]

Thus, the value of \( k \) is \( 1 \).