A circle having its centre in the first quadrant touches the y-axis at the point (0,2) and passes through the point (1,0). Find the equation of the circle

Answer

To find the equation of the circle that touches the y-axis at the point (0, 2) and passes through the point (1, 0), we can follow these steps:

### Step 1: Identify the center and radius of the circle
Since the circle touches the y-axis at (0, 2), the distance from the center of the circle to the y-axis is equal to the radius. Let the center of the circle be (H, K). Therefore, we have:
- The radius \( R = H \) (since it touches the y-axis).
- The point of tangency is (0, 2), which gives us \( K = 2 \).

### Step 2: Write the equation of the circle
The standard form of the equation of a circle is given by:
\[ (x - H)^2 + (y - K)^2 = R^2 \]
Substituting \( K = 2 \) and \( R = H \), we get:
\[ (x - H)^2 + (y - 2)^2 = H^2 \]

### Step 3: Substitute the point (1, 0) into the equation
Since the circle passes through the point (1, 0), we substitute \( x = 1 \) and \( y = 0 \) into the equation:
\[ (1 - H)^2 + (0 - 2)^2 = H^2 \]
This simplifies to:
\[ (1 - H)^2 + 4 = H^2 \]

### Step 4: Expand and simplify the equation
Expanding \( (1 - H)^2 \):
\[ 1 - 2H + H^2 + 4 = H^2 \]
This simplifies to:
\[ 5 - 2H + H^2 = H^2 \]
Subtracting \( H^2 \) from both sides:
\[ 5 - 2H = 0 \]

### Step 5: Solve for H
Rearranging gives:
\[ 2H = 5 \implies H = \frac{5}{2} \]

### Step 6: Write the final equation of the circle
Now that we have \( H = \frac{5}{2} \) and \( K = 2 \), we can substitute these values back into the equation of the circle:
\[ (x - \frac{5}{2})^2 + (y - 2)^2 = \left(\frac{5}{2}\right)^2 \]
Calculating \( \left(\frac{5}{2}\right)^2 \):
\[ \left(\frac{5}{2}\right)^2 = \frac{25}{4} \]
Thus, the equation of the circle is:
\[ \left(x - \frac{5}{2}\right)^2 + (y - 2)^2 = \frac{25}{4} \]

### Final Answer
The equation of the circle is:
\[ \left(x - \frac{5}{2}\right)^2 + (y - 2)^2 = \frac{25}{4} \]