点 $P(-2, 0)$，$Q(5, 1)$，$R(6, -6)$ 和 $S(-1, -7)$ 在同一个圆上。 $PQ$ 和 $SR$ 是长度相同的平行弦。 证明圆心位于 $(2, -3)$。 求该圆的方程。 第二个圆的圆心为 $C$。已知 $y$ 轴是 $CR$ 的垂直平分线，且也是第二个圆的切线。 写出 $C$ 的坐标。 由此解释为什么 $x$ 轴也是第二个圆的切线。

1. Identify the shape formed by the four points

Since $PQ$ and $SR$ are parallel chords of the same length, $PQRS$ is an isosceles trapezium or a rectangle. Since they lie on a circle, the parallel chords of equal length must form a rectangle (or square). The diagonals of a rectangle inscribed in a circle are diameters of the circle.

2. Calculate the centre of the circle using the midpoint of a diagonal

The centre of the circle is the midpoint of the diagonal $PR$ (or $QS$).
Midpoint of $PR = \left( \frac{-2 + 6}{2}, \frac{0 + (-6)}{2} \right)$
$= \left( \frac{4}{2}, \frac{-6}{2} \right)$
$= (2, -3)$

1. Identify the centre of the circle from part (a)

The centre of the circle is $(2, -3)$.

2. Calculate the square of the radius using the centre and a point on the circle

Using point $P(-2, 0)$:
$r^2 = (-2 - 2)^2 + (0 - (-3))^2$
$r^2 = (-4)^2 + (3)^2$
$r^2 = 16 + 9 = 25$

3. Write the equation of the circle

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

1. Find the coordinates of $C$ using the property of the perpendicular bisector

Since the $y$-axis is the perpendicular bisector of $CR$, $C$ is the reflection of $R(6, -6)$ across the $y$-axis.
The $x$-coordinate of $C$ is $-6$ and the $y$-coordinate remains $-6$.
$C = (-6, -6)$

1. Determine the radius of the second circle

Since the $y$-axis is a tangent to the second circle and the centre is $C(-6, -6)$, the radius of the circle is the perpendicular distance from $C$ to the $y$-axis.
Radius $= |-6| = 6$

2. Show that the distance to the $x$-axis equals the radius

The perpendicular distance from the centre $C(-6, -6)$ to the $x$-axis is $|-6| = 6$.
Since this distance is equal to the radius of the circle, the $x$-axis is also a tangent to the second circle.