A circle passes thorugh A(2,1) and touches y-axis then the locus of its centre is

Options

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

Answer

To find the locus of the center of a circle that passes through the point A(2, 1) and touches the y-axis, we can follow these steps:

### Step 1: Define the center of the circle
Let the center of the circle be at point (H, K).

### Step 2: Use the properties of the circle
Since the circle touches the y-axis, the distance from the center to the y-axis is equal to the radius of the circle. Therefore, the x-coordinate of the center (H) is equal to the radius (R) of the circle. Thus, we have:
\[ R = H \]

### Step 3: Write the general equation of the circle
The general equation of a circle with center (H, K) and radius R is given by:
\[ (X - H)^2 + (Y - K)^2 = R^2 \]
Substituting \( R \) with \( H \), we get:
\[ (X - H)^2 + (Y - K)^2 = H^2 \]

### Step 4: Substitute the point A(2, 1)
Since the circle passes through the point A(2, 1), we can substitute these coordinates into the circle's equation:
\[ (2 - H)^2 + (1 - K)^2 = H^2 \]

### Step 5: Expand the equation
Expanding the left-hand side:
\[ (2 - H)^2 = 4 - 4H + H^2 \]
\[ (1 - K)^2 = 1 - 2K + K^2 \]
So, we have:
\[ 4 - 4H + H^2 + 1 - 2K + K^2 = H^2 \]

### Step 6: Simplify the equation
Now, simplify the equation:
\[ 5 - 4H - 2K + K^2 = 0 \]
Rearranging gives:
\[ K^2 - 2K - 4H + 5 = 0 \]

### Step 7: Identify the locus
This is a quadratic equation in K. The locus of the center (H, K) can be represented as:
\[ K^2 - 2K + 5 = 4H \]
This represents a parabola in the K-H coordinate system.

### Final Equation
Thus, the locus of the center of the circle is given by:
\[ K^2 - 2K = 4H - 5 \]