圆 $C_1$ 的方程为 $x^2 + y^2 + 2x - 6y = 10$。 求该圆的圆心坐标和准确半径。 A、B、E 和 F 是另一个圆 $C_2$ 上的 4 个点。 $EF$ 是弦 $AB$ 的垂直平分线。$EF$ 与 $AB$ 交于点 $\left(-\dfrac{9}{2},\,\dfrac{11}{2}\right)$。 点 A 的坐标为 $(-2,\,8)$。 弦 $AB$ 的垂直平分线方程为 $y + x = 1$。直线 $2y = x + 8$ 也通过该圆的圆心。P(2, 0) 是该圆上的另一个点。 求 $B$ 的坐标。 解释为什么角 $EAF = 90^{\circ}$。 求该圆的圆心。 求圆在点 $P$ 处的切线方程。

1. Group the $x$ and $y$ terms and complete the square for both.

$x^2 + 2x + y^2 - 6y = 10$
$(x + 1)^2 - 1^2 + (y - 3)^2 - (-3)^2 = 10$

2. Rearrange the equation into the standard circle form $(x-a)^2 + (y-b)^2 = r^2$.

$(x + 1)^2 - 1 + (y - 3)^2 - 9 = 10$
$(x + 1)^2 + (y - 3)^2 = 20$

3. Identify the centre and calculate the exact radius.

Centre $= (-1, 3)$
Radius $r = \sqrt{20} = 2\sqrt{5}$

1. Identify the relationship between the given points

Since $EF$ is the perpendicular bisector of $AB$, the point where they intersect, $\left(-\frac{9}{2}, \frac{11}{2}\right)$, is the midpoint of $AB$.

2. Use the midpoint formula to find the coordinates of $B$

Let the coordinates of $B$ be $(x, y)$.
Using the midpoint formula:
$\frac{x + (-2)}{2} = -\frac{9}{2} \implies x - 2 = -9 \implies x = -7$
$\frac{y + 8}{2} = \frac{11}{2} \implies y + 8 = 11 \implies y = 3$

1. Identify that $EF$ is a diameter of the circle

The perpendicular bisector of any chord of a circle passes through the centre of the circle. Since $EF$ is the perpendicular bisector of chord $AB$ and its endpoints $E$ and $F$ lie on the circle, $EF$ is a diameter of $C_2$.

2. Use the angle in a semicircle property to conclude the proof

Since $EF$ is a diameter and $A$ is a point on the circumference of the circle, the angle subtended by $EF$ at $A$ is $90^{\circ}$ (angle in a semicircle). Therefore, $\angle EAF = 90^{\circ}$.

1. Set up simultaneous equations for the centre of the circle

The centre of the circle lies on the perpendicular bisector of any chord, so it lies on $y + x = 1$.
It is also given that the line $2y = x + 8$ passes through the centre.
We solve the simultaneous equations:
$y = 1 - x$ --- (1)
$2y = x + 8$ --- (2)

2. Solve the simultaneous equations to find the coordinates of the centre

Substitute (1) into (2):
$2(1 - x) = x + 8$
$2 - 2x = x + 8$
$3x = -6$
$x = -2$
Substitute $x = -2$ into (1):
$y = 1 - (-2) = 3$
The centre of the circle is $(-2, 3)$.

1. State the coordinates of the centre of the circle (found in part (iii))

From part (b)(iii), the centre of the circle $C$ is the intersection of $y + x = 1$ and $2y = x + 8$.
$2(1 - x) = x + 8 \implies 3x = -6 \implies x = -2$
$y = 1 - (-2) = 3$
Centre $C$ is $(-2, 3)$.

2. Find the gradient of the radius at $P$

Gradient of radius $CP = \frac{0 - 3}{2 - (-2)} = -\frac{3}{4}$

3. Find the gradient of the tangent and its equation

Since the tangent is perpendicular to the radius at the point of contact,
Gradient of tangent $= \frac{-1}{-\frac{3}{4}} = \frac{4}{3}$

Equation of the tangent at $P(2, 0)$:
$y - 0 = \frac{4}{3}(x - 2)$
$3y = 4x - 8$