圆 $C_1$ 的方程为 $x^2 + y^2 - 4x + 6y - 3 = 0$ 。

(i)
求 $C_1$ 的圆心坐标和半径。
(ii)
解释为什么 $C_1$ 与直线 $y = 1$ 相切。
(iii)
求圆 $C_2$ 的方程，该圆是 $C_1$ 关于直线 $y = 1$ 对称得到的。
(iv)
直线 $y + 9 = x$ 与圆 $C_1$ 交于点 $M(2, -7)$ 和 $N(6, -3)$ 。

求圆心到直线 $y + 9 = x$ 的最短距离。

圆的方程中点公式平面几何证明 English

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完整解法

9(i)

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

$x^2 - 4x + y^2 + 6y = 3$
$(x - 2)^2 - 2^2 + (y + 3)^2 - 3^2 = 3$
$(x - 2)^2 - 4 + (y + 3)^2 - 9 = 3$

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

$(x - 2)^2 + (y + 3)^2 = 3 + 4 + 9$
$(x - 2)^2 + (y + 3)^2 = 16$

3. Identify the centre and radius from the standard form.

Centre is $(2, -3)$ .
Radius $r = \sqrt{16} = 4$ .

答案

Centre $(2, -3)$ , Radius $4$

9(ii)

1. Calculate the perpendicular distance from the centre of $C_1$ to the line $y = 1$.

From (i), the centre of $C_1$ is $(2, -3)$ and the radius is $4$ .
The perpendicular distance from the centre $(2, -3)$ to the horizontal line $y = 1$ is:
$|1 - (-3)| = 4$

2. Conclude why the line touches the circle.

Since the distance from the centre to the line is exactly equal to the radius ( $4$ ), the line is a tangent to the circle and therefore touches it.

答案

The distance from the centre of the circle to the line $y = 1$ is equal to the radius of the circle, hence the line touches the circle.

9(iii)

1. Find the coordinates of the centre of $C_2$ by reflecting the centre of $C_1$ in the line $y = 1$

From part (i), the centre of $C_1$ is $(2, -3)$ .
The line of reflection is $y = 1$ .
The $x$ -coordinate remains $2$ .
The distance from the $y$ -coordinate $-3$ to the line $y = 1$ is $1 - (-3) = 4$ .
The $y$ -coordinate of the reflected centre is $1 + 4 = 5$ .
Centre of $C_2$ is $(2, 5)$ .

2. Write the equation of $C_2$

The radius of $C_2$ is the same as $C_1$ , which is $4$ .
The equation of $C_2$ is:
$(x - 2)^2 + (y - 5)^2 = 4^2$
$(x - 2)^2 + (y - 5)^2 = 16$

答案

$(x - 2)^2 + (y - 5)^2 = 16$

9(iv)

1. Find the midpoint of the chord $MN$

Let the midpoint of $MN$ be $P$ .
$P = \left( \frac{2+6}{2}, \frac{-7+(-3)}{2} \right) = (4, -5)$

2. Calculate the distance from the centre of the circle to the midpoint of the chord

From part (i), the centre of $C_1$ is $(2, -3)$ .
The shortest distance is the perpendicular distance from the centre to the chord, which passes through the midpoint of the chord.
Distance $= \sqrt{(4-2)^2 + (-5 - (-3))^2}$
$= \sqrt{2^2 + (-2)^2}$
$= \sqrt{8} = 2\sqrt{2}$

答案

$2\sqrt{2}$

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圆 C1C_1C1​ 的方程为 x2+y2−4x+6y−3=0x^2 + y^2 - 4x + 6y - 3 = 0x2+y2−4x+6y−3=0。