Find the equations of the following circles :
(i) centre (0,2) and radius 2
(ii) Centre `((1)/(2), (1)/(4))` and radius `(1)/(12)`
(iii) centre (1,1) and radius `sqrt(2)`.

Answer

To find the equations of the given circles, we will use the standard equation of a circle, which is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Part (i):
**Given:** Center \((0, 2)\) and radius \(2\).

1. Substitute \(h = 0\), \(k = 2\), and \(r = 2\) into the equation:
\[ (x - 0)^2 + (y - 2)^2 = 2^2 \]
2. This simplifies to:
\[ x^2 + (y - 2)^2 = 4 \]
3. Now, expand \((y - 2)^2\):
\[ x^2 + (y^2 - 4y + 4) = 4 \]
4. Combine like terms:
\[ x^2 + y^2 - 4y + 4 - 4 = 0 \]
5. Thus, the equation of the circle is:
\[ x^2 + y^2 - 4y = 0 \]

### Part (ii):
**Given:** Center \(\left(\frac{1}{2}, \frac{1}{4}\right)\) and radius \(\frac{1}{12}\).

1. Substitute \(h = \frac{1}{2}\), \(k = \frac{1}{4}\), and \(r = \frac{1}{12}\) into the equation:
\[ \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \left(\frac{1}{12}\right)^2 \]
2. This simplifies to:
\[ \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{144} \]

### Part (iii):
**Given:** Center \((1, 1)\) and radius \(\sqrt{2}\).

1. Substitute \(h = 1\), \(k = 1\), and \(r = \sqrt{2}\) into the equation:
\[ (x - 1)^2 + (y - 1)^2 = (\sqrt{2})^2 \]
2. This simplifies to:
\[ (x - 1)^2 + (y - 1)^2 = 2 \]

### Summary of the Equations:
1. For part (i): \(x^2 + y^2 - 4y = 0\)
2. For part (ii): \(\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{144}\)
3. For part (iii): \((x - 1)^2 + (y - 1)^2 = 2\)