题目

Problem

Figure 2 shows a sketch of

Given that $l$ is the tangent to $C$ at the point $N$ ,

(a) show that an equation for the straight line passing through $X$ and $N$ is $2x + 5y + 7 = 0$ .

(b) Hence find

(i) the coordinates of $N$ ,

(ii) an equation for $C$ .

(8)

题目中文翻译

图2显示了以下图形的草图：

已知 $l$ 是 $C$ 在点 $N$ 处的切线，

(a) 证明过 $X$ 和 $N$ 的直线方程为 $2x + 5y + 7 = 0$ 。

(b) 由此求

(i) $N$ 的坐标，

(ii) $C$ 的方程。

（8分）

解答

切线与半径垂直：切线 $l$ 的斜率为 $\dfrac{2}{5}$ ，因此 $XN$ 的斜率为 $-\dfrac{5}{2}$ 。利用点斜式求 $XN$ 的方程。

(a) The line $l$ has gradient $\dfrac{2}{5}$ .

Since $l$ is the tangent to $C$ at $N$ , the radius $XN$ is perpendicular to $l$ . Therefore the gradient of $XN$ is $-\dfrac{5}{2}$ .

Using the point $X(4, -3)$ with gradient $-\dfrac{5}{2}$ :

y - (-3) =&\, -\frac{5}{2}(x - 4) \\[2mm] y + 3 =&\, -\frac{5}{2}x + 10 \\[2mm] 2(y + 3) =&\, -5(x - 4) \\ 2y + 6 =&\, -5x + 20 \\ 2x + 5y + 7 =&\, 0 \end{align*}$$ $$\boxed{2x + 5y + 7 = 0}$$ **(b)(i)** Solving $2x + 5y + 7 = 0$ simultaneously with the equation of line $l$: $$\begin{align*} x =&\, 9, \quad y = -5 \end{align*}$$ $$\boxed{N = (9, -5)}$$ **(b)(ii)** The circle $C$ has centre $X(4, -3)$ and radius $|XN|$: $$\begin{align*} r^2 =&\, (9 - 4)^2 + (-5 - (-3))^2 \\ =&\, 25 + 4 \\ =&\, 29 \end{align*}$$ The equation of circle $C$ is: $$\boxed{(x - 4)^2 + (y + 3)^2 = 29}$$