The equation of a circle which has its centre on the positive side of x-axis and cuts off a chord of length 2 along the line `sqrt(3)y-x=0` and also touches the line y = x is

Options

A. `x^(2)+y^(2)-4x+1=0`
B. `x^(2)+y^(2)-4x+2=0`
C. `x^(2)+y^(2)-8x+8=0`
D. `x^(2)+y^(2)-8x+4=0`

Answer

To find the equation of the circle that meets the specified conditions, we can follow these steps:

### Step 1: Define the center of the circle
Let the center of the circle be \( C(a, 0) \) since it lies on the positive x-axis.

### Step 2: Determine the distance from the center to the line \( \sqrt{3}y - x = 0 \)
The distance \( d \) from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \) is given by the formula:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( \sqrt{3}y - x = 0 \), we can rewrite it in the form \( -x + \sqrt{3}y + 0 = 0 \) where \( A = -1, B = \sqrt{3}, C = 0 \). Thus, the distance from \( C(a, 0) \) to the line is:
\[ d = \frac{|-a + 0|}{\sqrt{(-1)^2 + (\sqrt{3})^2}} = \frac{a}{2} \]

### Step 3: Find the midpoint of the chord
Since the chord length is 2, the distance from the center \( C \) to the midpoint \( M \) of the chord is:
\[ CM = \frac{1}{2} \text{ (half of the chord length)} \]
Thus, we have:
\[ CM = 1 \]

### Step 4: Relate the distances using Pythagoras theorem
In triangle \( ACM \), where \( A \) is one endpoint of the chord, we have:
\[ AC^2 = AM^2 + CM^2 \]
Here, \( AC \) is the radius \( r \) of the circle, \( AM = 1 \), and \( CM = \frac{a}{2} \):
\[ r^2 = 1^2 + \left(\frac{a}{2}\right)^2 \]
This simplifies to:
\[ r^2 = 1 + \frac{a^2}{4} \]

### Step 5: Find the distance from the center to the line \( y = x \)
The distance from the center \( C(a, 0) \) to the line \( y = x \) can be calculated similarly:
\[ d = \frac{|a - 0|}{\sqrt{1^2 + (-1)^2}} = \frac{a}{\sqrt{2}} \]
Since the circle touches the line \( y = x \), this distance is equal to the radius \( r \):
\[ r = \frac{a}{\sqrt{2}} \]

### Step 6: Set up the equations
From steps 4 and 5, we have two expressions for \( r^2 \):
1. \( r^2 = 1 + \frac{a^2}{4} \)
2. \( r^2 = \left(\frac{a}{\sqrt{2}}\right)^2 = \frac{a^2}{2} \)

Setting these equal gives:
\[ 1 + \frac{a^2}{4} = \frac{a^2}{2} \]

### Step 7: Solve for \( a \)
Rearranging the equation:
\[ 1 = \frac{a^2}{2} - \frac{a^2}{4} \]
\[ 1 = \frac{2a^2 - a^2}{4} \implies 1 = \frac{a^2}{4} \implies a^2 = 4 \implies a = 2 \]

### Step 8: Find the radius
Substituting \( a = 2 \) back into the equation for \( r \):
\[ r = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 9: Write the equation of the circle
The equation of the circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 2, k = 0, r = \sqrt{2} \):
\[ (x - 2)^2 + (y - 0)^2 = (\sqrt{2})^2 \]
This simplifies to:
\[ (x - 2)^2 + y^2 = 2 \]

### Step 10: Expand the equation
Expanding the equation gives:
\[ x^2 - 4x + 4 + y^2 = 2 \implies x^2 + y^2 - 4x + 2 = 0 \]

Thus, the final equation of the circle is:
\[ \boxed{x^2 + y^2 - 4x + 2 = 0} \]