The equation to a circle with centre(2,1) and touching x axis is

Options

A. `x^2+y^2+4x+2y+4=0`
B. `x^2+y^2-4x-2y+4=0`
C. `x^2+y^2-4x-2y+1=0`
D. none of these

Answer

To find the equation of a circle with center (2, 1) that touches the x-axis, we can follow these steps:

### Step 1: Identify the center and radius
The center of the circle is given as (2, 1). Since the circle touches the x-axis, the distance from the center to the x-axis will be equal to the radius of the circle. The y-coordinate of the center is 1, which means the radius \( r \) is 1.

### Step 2: Use the standard equation of a circle
The standard equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, \( h = 2 \), \( k = 1 \), and \( r = 1 \).

### Step 3: Substitute the values into the equation
Substituting the values into the standard equation:
\[ (x - 2)^2 + (y - 1)^2 = 1^2 \]
This simplifies to:
\[ (x - 2)^2 + (y - 1)^2 = 1 \]

### Step 4: Expand the equation
Now, we will expand the equation:
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ (y - 1)^2 = y^2 - 2y + 1 \]
Combining these, we get:
\[ x^2 - 4x + 4 + y^2 - 2y + 1 = 1 \]

### Step 5: Simplify the equation
Now, we simplify the equation:
\[ x^2 + y^2 - 4x - 2y + 5 = 1 \]
Subtracting 1 from both sides gives:
\[ x^2 + y^2 - 4x - 2y + 4 = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ x^2 + y^2 - 4x - 2y + 4 = 0 \]

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