The locus of centre of the circle touching x-axis nad the line `y=x` is

Options

A. `y=(sqrt(2)-1)x`
B. `x=(sqrt(2)+1)y`
C. `y=2x`
D. `y=-x`

Answer

To find the locus of the center of a circle that touches the x-axis and the line \( y = x \), we can follow these steps:

### Step 1: Understand the properties of the circle
A circle that touches the x-axis has its center at a point \( (h, k) \) where the radius \( r \) is equal to the y-coordinate of the center. Therefore, we can say:
\[ r = k \]

### Step 2: Use the condition for touching the line \( y = x \)
The distance from the center of the circle \( (h, k) \) to the line \( y = x \) must also be equal to the radius \( r \). The equation of the line \( y = x \) can be rewritten in the form \( Ax + By + C = 0 \) as:
\[ x - y = 0 \]
Here, \( A = 1, B = -1, C = 0 \).

The formula for the distance \( d \) from a point \( (h, k) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} = \frac{|h - k|}{\sqrt{1^2 + (-1)^2}} = \frac{|h - k|}{\sqrt{2}} \]

### Step 3: Set the distance equal to the radius
Since the circle touches the line \( y = x \), we set the distance equal to the radius:
\[ \frac{|h - k|}{\sqrt{2}} = k \]

### Step 4: Solve for \( h \)
Multiplying both sides by \( \sqrt{2} \) gives:
\[ |h - k| = k\sqrt{2} \]
This leads to two cases:
1. \( h - k = k\sqrt{2} \)
2. \( h - k = -k\sqrt{2} \)

### Step 5: Solve each case
**Case 1:**
\[ h - k = k\sqrt{2} \implies h = k\sqrt{2} + k = k(\sqrt{2} + 1) \]

**Case 2:**
\[ h - k = -k\sqrt{2} \implies h = -k\sqrt{2} + k = k(1 - \sqrt{2}) \]

### Step 6: Express \( h \) in terms of \( k \)
From both cases, we can express \( h \) in terms of \( k \):
1. \( h = k(\sqrt{2} + 1) \)
2. \( h = k(1 - \sqrt{2}) \)

### Step 7: Identify the locus
To find the locus, we can express \( k \) in terms of \( h \):
1. From \( h = k(\sqrt{2} + 1) \), we can write:
\[ k = \frac{h}{\sqrt{2} + 1} \]
Thus, the equation becomes:
\[ y = \frac{x}{\sqrt{2} + 1} \]

2. From \( h = k(1 - \sqrt{2}) \), we can write:
\[ k = \frac{h}{1 - \sqrt{2}} \]
Thus, the equation becomes:
\[ y = \frac{x}{1 - \sqrt{2}} \]

### Conclusion
The locus of the center of the circle is given by the equations derived above, which can be simplified to find the relationship between \( x \) and \( y \).