圆 $C_1$ 的方程由 $x^2 + y^2 - 4x - 8y = 0$ 给出。

(a)
求 $C_1$ 的圆心坐标和半径。
(b)
点 $A\ (0,8)$ 在圆 $C_1$ 上。

求圆 $C_1$ 在点 $A$ 处的切线方程。
(c)
将圆 $C_1$ 关于 $y$ 轴对称。

求表示对称后的圆 $C_2$ 的方程。
(d)
圆 $C_3$ 的方程为 $(x-5)^2 + (y-8)^2 = 5$ 。

说明 $C_1$ 和 $C_3$ 是否相交。

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完整解法

9(a)

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

$x^2 - 4x + y^2 - 8y = 0$
$(x - 2)^2 - 2^2 + (y - 4)^2 - 4^2 = 0$
$(x - 2)^2 - 4 + (y - 4)^2 - 16 = 0$
$(x - 2)^2 + (y - 4)^2 = 20$

2. Identify the center and radius from the standard form $(x-a)^2 + (y-b)^2 = r^2$

Center $(a, b) = (2, 4)$
Radius $r = \sqrt{20} = 2\sqrt{5}$

答案

Center: $(2, 4)$ , Radius: $\sqrt{20}$ or $2\sqrt{5}$

9(b)

1. Calculate the gradient of the radius connecting the center to point $A$.

From part (a), the center of $C_1$ is $(2, 4)$ .
Gradient of radius at $A(0, 8) = \frac{8 - 4}{0 - 2} = \frac{4}{-2} = -2$

2. Determine the gradient of the tangent at $A$.

Since the tangent is perpendicular to the radius at the point of contact:
Gradient of tangent $= -\frac{1}{-2} = \frac{1}{2}$

3. Form the equation of the tangent line.

Using the point-gradient form with $A(0, 8)$ :
$y - 8 = \frac{1}{2}(x - 0)$
$y = \frac{1}{2}x + 8$

答案

$y = \frac{1}{2}x + 8$

9(c)

1. Find the center of the reflected circle $C_2$

From part (a), the center of $C_1$ is $(2, 4)$ and the radius squared is $20$ .
When reflected about the $y$ -axis, the $x$ -coordinate of the center changes sign.
Center of $C_2 = (-2, 4)$

2. Write the equation of the reflected circle $C_2$

The radius remains the same.
Equation of $C_2$ : $(x - (-2))^2 + (y - 4)^2 = 20$
$(x + 2)^2 + (y - 4)^2 = 20$
Expanding this gives:
$x^2 + 4x + 4 + y^2 - 8y + 16 = 20$
$x^2 + y^2 + 4x - 8y = 0$

答案

$x^2 + y^2 + 4x - 8y = 0$ or $(x+2)^2 + (y-4)^2 = 20$

9(d)

1. State the centre and radius of $C_1$ and $C_3$.

From part (a), for $C_1$ :
Centre $O_1 = (2, 4)$
Radius $r_1 = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}$

For $C_3$ :
Centre $O_3 = (5, 8)$
Radius $r_3 = \sqrt{5}$

2. Calculate the distance between the centres of the two circles.

Distance $O_1O_3 = \sqrt{(5 - 2)^2 + (8 - 4)^2}$
$O_1O_3 = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

3. Calculate the sum and difference of the radii.

Sum of radii, $r_1 + r_3 = 2\sqrt{5} + \sqrt{5} = 3\sqrt{5} \approx 6.71$
Difference of radii, $r_1 - r_3 = 2\sqrt{5} - \sqrt{5} = \sqrt{5} \approx 2.24$

4. Compare the distance between centres with the sum and difference of radii to determine intersection.

Since $r_1 - r_3 < O_1O_3 < r_1 + r_3$ (i.e., $2.24 < 5 < 6.71$ ), the circles intersect each other at two points.

答案

Since the distance between the centres ( $5$ ) is less than the sum of their radii ( $\approx 6.71$ ) and greater than the difference of their radii ( $\approx 2.24$ ), $C_1$ and $C_3$ intersect each other.

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圆 C1C_1C1​ 的方程由 x2+y2−4x−8y=0x^2 + y^2 - 4x - 8y = 0x2+y2−4x−8y=0 给出。