Find the centre and radius of each of the following circle :
(i) `x^(2)+y^(2)+4x-4y+1=0`
(ii) `2x^(2)+2y^(2)-6x+6y+1=0`
(iii) `2x^(2)+2y^(2)=3k(x+k)`
(iv) `3x^(2)+3y^(2)-5x-6y+4=0`
(v) `x^(2)+y^(2)-2ax-2by+a^(2)=0`

Answer

To find the center and radius of each of the given circles, we will use the standard form of the equation of a circle, which is:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From this equation:
- The center of the circle is given by the coordinates \((-g, -f)\).
- The radius of the circle is given by \(r = \sqrt{g^2 + f^2 - c}\).

Now, let's solve each part step by step.

### (i) \(x^2 + y^2 + 4x - 4y + 1 = 0\)

1. Identify coefficients:
- Coefficient of \(x\) is \(4\) (so \(2g = 4\) → \(g = 2\))
- Coefficient of \(y\) is \(-4\) (so \(2f = -4\) → \(f = -2\))
- Constant term \(c = 1\)

2. Find the center:
\[ \text{Center} = (-g, -f) = (-2, 2) \]

3. Find the radius:
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{2^2 + (-2)^2 - 1} = \sqrt{4 + 4 - 1} = \sqrt{7} \]

### (ii) \(2x^2 + 2y^2 - 6x + 6y + 1 = 0\)

1. Divide the entire equation by \(2\):
\[ x^2 + y^2 - 3x + 3y + \frac{1}{2} = 0 \]

2. Identify coefficients:
- Coefficient of \(x\) is \(-3\) (so \(2g = -3\) → \(g = -\frac{3}{2}\))
- Coefficient of \(y\) is \(3\) (so \(2f = 3\) → \(f = \frac{3}{2}\))
- Constant term \(c = \frac{1}{2}\)

3. Find the center:
\[ \text{Center} = \left(-\left(-\frac{3}{2}\right), -\frac{3}{2}\right) = \left(\frac{3}{4}, -\frac{3}{4}\right) \]

4. Find the radius:
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right)^2 - \frac{1}{2}} = \sqrt{\frac{9}{4} + \frac{9}{4} - \frac{1}{2}} = \sqrt{\frac{18}{4} - \frac{2}{4}} = \sqrt{\frac{16}{4}} = 2 \]

### (iii) \(2x^2 + 2y^2 = 3kx + k\)

1. Rearranging gives:
\[ 2x^2 + 2y^2 - 3kx - k = 0 \]

2. Divide the entire equation by \(2\):
\[ x^2 + y^2 - \frac{3k}{2}x - \frac{k}{2} = 0 \]

3. Identify coefficients:
- Coefficient of \(x\) is \(-\frac{3k}{2}\) (so \(g = -\frac{3k}{4}\))
- Coefficient of \(y\) is \(0\) (so \(f = 0\))
- Constant term \(c = -\frac{k}{2}\)

4. Find the center:
\[ \text{Center} = \left(-\left(-\frac{3k}{4}\right), 0\right) = \left(\frac{3k}{4}, 0\right) \]

5. Find the radius:
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{3k}{4}\right)^2 + 0^2 - \left(-\frac{k}{2}\right)} = \sqrt{\frac{9k^2}{16} + \frac{k}{2}} = \sqrt{\frac{9k^2 + 8k}{16}} = \frac{\sqrt{9k^2 + 8k}}{4} \]

### (iv) \(3x^2 + 3y^2 - 5x - 6y + 4 = 0\)

1. Divide the entire equation by \(3\):
\[ x^2 + y^2 - \frac{5}{3}x - 2y + \frac{4}{3} = 0 \]

2. Identify coefficients:
- Coefficient of \(x\) is \(-\frac{5}{3}\) (so \(g = -\frac{5}{6}\))
- Coefficient of \(y\) is \(-2\) (so \(f = -1\))
- Constant term \(c = \frac{4}{3}\)

3. Find the center:
\[ \text{Center} = \left(-\left(-\frac{5}{6}\right), -(-1)\right) = \left(\frac{5}{6}, 1\right) \]

4. Find the radius:
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{5}{6}\right)^2 + (-1)^2 - \frac{4}{3}} = \sqrt{\frac{25}{36} + 1 - \frac{4}{3}} = \sqrt{\frac{25}{36} + \frac{36}{36} - \frac{48}{36}} = \sqrt{\frac{13}{36}} = \frac{\sqrt{13}}{6} \]

### (v) \(x^2 + y^2 - 2ax - 2by + a^2 = 0\)

1. Identify coefficients:
- Coefficient of \(x\) is \(-2a\) (so \(g = -a\))
- Coefficient of \(y\) is \(-2b\) (so \(f = -b\))
- Constant term \(c = a^2\)

2. Find the center:
\[ \text{Center} = (a, b) \]

3. Find the radius:
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{(-a)^2 + (-b)^2 - a^2} = \sqrt{a^2 + b^2 - a^2} = \sqrt{b^2} = |b| \]

### Summary of Results:
- (i) Center: \((-2, 2)\), Radius: \(\sqrt{7}\)
- (ii) Center: \(\left(\frac{3}{4}, -\frac{3}{4}\right)\), Radius: \(2\)
- (iii) Center: \(\left(\frac{3k}{4}, 0\right)\), Radius: \(\frac{\sqrt{9k^2 + 8k}}{4}\)
- (iv) Center: \(\left(\frac{5}{6}, 1\right)\), Radius: \(\frac{\sqrt{13}}{6}\)
- (v) Center: \((a, b)\), Radius: \(|b|\)