Find the equations of the circles the end points of whose diameter are as follows :
(i) (0,6) and (6,0)
(ii) (-1,3) and (2,4)
(iii) (A+b,a-b) and (a-b,a+b)
(iv) (0,0) and (2,2)

Answer

To find the equations of the circles whose diameters are defined by the given endpoints, we can use the general equation of a circle derived from the endpoints of the diameter. The equation of the circle can be expressed as:

\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]

Where \((x_1, y_1)\) and \((x_2, y_2)\) are the endpoints of the diameter.

Let's solve each part step by step.

### (i) Endpoints: (0, 6) and (6, 0)

1. Identify the endpoints:
\[ (x_1, y_1) = (0, 6), \quad (x_2, y_2) = (6, 0) \]

2. Substitute into the circle equation:
\[ (x - 0)(x - 6) + (y - 6)(y - 0) = 0 \]

3. Simplify:
\[ x(x - 6) + (y - 6)y = 0 \]
\[ x^2 - 6x + y^2 - 6y = 0 \]

4. Rearranging gives:
\[ x^2 + y^2 - 6x - 6y = 0 \]

### (ii) Endpoints: (-1, 3) and (2, 4)

1. Identify the endpoints:
\[ (x_1, y_1) = (-1, 3), \quad (x_2, y_2) = (2, 4) \]

2. Substitute into the circle equation:
\[ (x + 1)(x - 2) + (y - 3)(y - 4) = 0 \]

3. Simplify:
\[ (x^2 - 2x + x - 2) + (y^2 - 4y + 3y - 12) = 0 \]
\[ x^2 - x - 2 + y^2 - y - 12 = 0 \]

4. Rearranging gives:
\[ x^2 + y^2 - x - y - 14 = 0 \]

### (iii) Endpoints: (A + b, a - b) and (a - b, a + b)

1. Identify the endpoints:
\[ (x_1, y_1) = (A + b, a - b), \quad (x_2, y_2) = (a - b, a + b) \]

2. Substitute into the circle equation:
\[ (x - (A + b))(x - (a - b)) + (y - (a - b))(y - (a + b)) = 0 \]

3. Expand:
\[ (x - (A + b))(x - (a - b)) + (y - (a - b))(y - (a + b)) = 0 \]
\[ (x^2 - (A + b + a - b)x + (A + b)(a - b)) + (y^2 - (2a)y + (a^2 - b^2)) = 0 \]

4. Combine and simplify:
\[ x^2 + y^2 - (A + b + a - b + 2a)y + (A + b)(a - b) + (a^2 - b^2) = 0 \]

### (iv) Endpoints: (0, 0) and (2, 2)

1. Identify the endpoints:
\[ (x_1, y_1) = (0, 0), \quad (x_2, y_2) = (2, 2) \]

2. Substitute into the circle equation:
\[ (x - 0)(x - 2) + (y - 0)(y - 2) = 0 \]

3. Simplify:
\[ x(x - 2) + y(y - 2) = 0 \]
\[ x^2 - 2x + y^2 - 2y = 0 \]

4. Rearranging gives:
\[ x^2 + y^2 - 2x - 2y = 0 \]

### Final Equations

1. (i) \(x^2 + y^2 - 6x - 6y = 0\)
2. (ii) \(x^2 + y^2 - x - y - 14 = 0\)
3. (iii) \(x^2 + y^2 - (A + b + a - b + 2a)y + (A + b)(a - b) + (a^2 - b^2) = 0\)
4. (iv) \(x^2 + y^2 - 2x - 2y = 0\)