一圆心为 O 的圆穿过点 P(-1,7) 和 Q(0,8)。

1. State the geometric relationship between a chord's perpendicular bisector and the circle's centre

Since $P$ and $Q$ lie on the circle, $PQ$ is a chord. The perpendicular bisector of any chord of a circle passes through its centre $O$.

1. Find the midpoint of the chord $PQ$

Midpoint of $PQ = \left(\frac{-1 + 0}{2}, \frac{7 + 8}{2}\right) = (-0.5, 7.5)$

2. Find the gradient of $PQ$ and the gradient of its perpendicular bisector

Gradient of $PQ = \frac{8 - 7}{0 - (-1)} = 1$
Gradient of perpendicular bisector $= -\frac{1}{1} = -1$

3. Find the equation of the perpendicular bisector of $PQ$

$y - 7.5 = -1(x - (-0.5))$
$y - 7.5 = -x - 0.5$
$y = -x + 7$

4. Solve simultaneously with the given line to find the $x$-coordinate of $O$

Since $O$ lies on both the perpendicular bisector and $y = 2x - 2$:
$-x + 7 = 2x - 2$
$3x = 9$
$x = 3$

5. Find the $y$-coordinate of $O$

$y = 2(3) - 2 = 4$
The coordinates of $O$ are $(3, 4)$.

1. Calculate the square of the radius using the centre O(3, 4) from part (b) and point Q(0, 8).

From part (b), the centre is $O(3, 4)$.
Using point $Q(0, 8)$:
$r^2 = (0 - 3)^2 + (8 - 4)^2$
$r^2 = (-3)^2 + 4^2 = 9 + 16 = 25$

2. Write the equation of the circle.

The equation of the circle is $(x - a)^2 + (y - b)^2 = r^2$.
$(x - 3)^2 + (y - 4)^2 = 25$