Find the equation of the circle which passes through the points P(l, 0), Q(3, 0), and R(0, 2). Find also (i) the coordinates of the other point in which the axis of y cuts the circle, (n) the coordinates of the other end of the diameter through Q.

Answer

To find the equation of the circle that passes through the points P(1, 0), Q(3, 0), and R(0, 2), we will use the general equation of a circle and substitute the coordinates of the given points into it.

### Step-by-Step Solution:

1. **General Equation of a Circle**:
The standard equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \(g\), \(f\), and \(c\) are constants.

2. **Substituting Point P(1, 0)**:
Substitute \(x = 1\) and \(y = 0\) into the circle equation:
\[ 1^2 + 0^2 + 2g(1) + 2f(0) + c = 0 \]
Simplifying this gives:
\[ 1 + 2g + c = 0 \quad \text{(Equation 1)} \]

3. **Substituting Point Q(3, 0)**:
Substitute \(x = 3\) and \(y = 0\):
\[ 3^2 + 0^2 + 2g(3) + 2f(0) + c = 0 \]
Simplifying this gives:
\[ 9 + 6g + c = 0 \quad \text{(Equation 2)} \]

4. **Substituting Point R(0, 2)**:
Substitute \(x = 0\) and \(y = 2\):
\[ 0^2 + 2^2 + 2g(0) + 2f(2) + c = 0 \]
Simplifying this gives:
\[ 4 + 4f + c = 0 \quad \text{(Equation 3)} \]

5. **Solving the System of Equations**:
We now have three equations:
- From Equation 1: \(c = -1 - 2g\)
- From Equation 2: \(c = -9 - 6g\)
- From Equation 3: \(c = -4 - 4f\)

Set the equations for \(c\) equal to each other:
\[ -1 - 2g = -9 - 6g \]
Rearranging gives:
\[ 4g = -8 \implies g = -2 \]

Now substitute \(g = -2\) into Equation 1 to find \(c\):
\[ c = -1 - 2(-2) = -1 + 4 = 3 \]

Substitute \(g = -2\) into Equation 3 to find \(f\):
\[ 3 = -4 - 4f \implies 4f = -7 \implies f = -\frac{7}{4} \]

6. **Final Equation of the Circle**:
Substitute \(g\), \(f\), and \(c\) back into the general equation:
\[ x^2 + y^2 - 4x - \frac{7}{2}y + 3 = 0 \]
To eliminate the fraction, multiply through by 2:
\[ 2x^2 + 2y^2 - 8x - 7y + 6 = 0 \]

### Answers to Additional Parts:
(i) **Finding the other point where the y-axis cuts the circle**:
Set \(x = 0\) in the circle's equation:
\[ 2(0)^2 + 2y^2 - 8(0) - 7y + 6 = 0 \implies 2y^2 - 7y + 6 = 0 \]
Solving this quadratic equation:
\[ y = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 6}}{2 \cdot 2} = \frac{7 \pm \sqrt{49 - 48}}{4} = \frac{7 \pm 1}{4} \]
This gives \(y = 2\) (point R) and \(y = \frac{3}{2}\). Thus, the other point is \((0, \frac{3}{2})\).

(ii) **Finding the other end of the diameter through Q(3, 0)**:
The center of the circle can be found from the values of \(g\) and \(f\):
\[ \text{Center} = (-g, -f) = (2, \frac{7}{4}) \]
The diameter's other end can be found using the midpoint formula. The midpoint between Q(3, 0) and the other end (let's call it \(A\)) is the center:
\[ \left(\frac{3 + x_A}{2}, \frac{0 + y_A}{2}\right) = \left(2, \frac{7}{4}\right) \]
Solving gives:
\[ \frac{3 + x_A}{2} = 2 \implies 3 + x_A = 4 \implies x_A = 1 \]
\[ \frac{0 + y_A}{2} = \frac{7}{4} \implies y_A = \frac{7}{2} \]
Thus, the other end of the diameter is \((1, \frac{7}{2})\).

### Summary of Results:
- The equation of the circle is:
\[ 2x^2 + 2y^2 - 8x - 7y + 6 = 0 \]
- The coordinates of the other point where the y-axis cuts the circle are \((0, \frac{3}{2})\).
- The coordinates of the other end of the diameter through Q are \((1, \frac{7}{2})\).