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P2 2023 May Q3

A Level / Edexcel / P2

IAL 2023 May Paper · Question 3

题目

Problem

A circle CC has centre (2, 5)(2,\, 5).

Given that the point P(8, −3)P(8,\, -3) lies on CC,

(a) (i) find the radius of CC

(ii) find an equation for CC

(3)

(b) Find the equation of the tangent to CC at PP giving your answer in the form ax+by+c=0ax + by + c = 0 where aa, bb and cc are integers to be found.

(4)
题目中文翻译

圆 CC 的圆心为 (2,5)(2,5)。已知点 P(8,−3)P(8,-3) 在圆上，

(a) (i) 求 CC 的半径；(ii) 求 CC 的一个方程。（3 分）

(b) 求圆在 PP 处的切线方程，写成 ax+by+c=0ax+by+c=0 的形式，其中 a,b,ca,b,c 为待求的整数。（4 分）

解答

(a)

解法一

思路

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半径为圆心到 PP 的距离。由圆心 (2,5)(2,5) 与半径 10 写出圆的标准方程。

答题过程

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(i)

The radius is the distance from the centre to PP:

r= (8−2)2+(−3−5)2= 36+64= 100= 10\begin{align*} r =&\, \sqrt{(8 - 2)^2 + (-3 - 5)^2} \\ =&\, \sqrt{36 + 64} \\ =&\, \sqrt{100} \\ =&\, 10 \end{align*} r=10\boxed{r = 10}

(ii)

The equation of circle CC:

(x−2)2+(y−5)2=100\boxed{(x - 2)^2 + (y - 5)^2 = 100}

(b)

解法一

思路

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切线垂直半径，先求圆心到 PP 的斜率，取负倒数，再用点斜式整理为整数系数形式。

答题过程

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The gradient of CPCP:

mCP= −3−58−2= −86= −43\begin{align*} m_{CP} =&\, \frac{-3 - 5}{8 - 2} \\ =&\, \frac{-8}{6} \\ =&\, -\frac{4}{3} \end{align*}

Since the tangent at PP is perpendicular to CPCP:

mtangent=34m_{\text{tangent}} = \frac{3}{4}

Using point–slope form at P(8,−3)P(8, -3):

y−(−3)= 34(x−8)4(y+3)= 3(x−8)4y+12= 3x−243x−4y−36= 0\begin{align*} y - (-3) =&\, \frac{3}{4}(x - 8) \\ 4(y + 3) =&\, 3(x - 8) \\ 4y + 12 =&\, 3x - 24 \\ 3x - 4y - 36 =&\, 0 \end{align*} 3x−4y−36=0\boxed{3x - 4y - 36 = 0}

解法二

思路

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隐函数求导法。对圆的方程两边关于 xx 求导，代入 PP 点坐标求切线斜率，再用点斜式。

答题过程

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The circle equation is (x−2)2+(y−5)2=100(x - 2)^2 + (y - 5)^2 = 100.

Differentiating implicitly with respect to xx:

2(x−2)+2(y−5)dydx=02(x - 2) + 2(y - 5)\frac{\mathrm{d}y}{\mathrm{d}x} = 0 dydx=−x−2y−5\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{x - 2}{y - 5}

At P(8,−3)P(8, -3):

dydx= −8−2−3−5= −6−8= 34\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\, -\frac{8 - 2}{-3 - 5} \\ =&\, -\frac{6}{-8} \\ =&\, \frac{3}{4} \end{align*}

Using point–slope form at P(8,−3)P(8, -3):

y+3=34(x−8)y + 3 = \frac{3}{4}(x - 8) 3x−4y−36=0\boxed{3x - 4y - 36 = 0}

解法三

思路

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切线公式法。圆 (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2 上点 (x1,y1)(x_1,y_1) 处的切线方程为 (x1−a)(x−a)+(y1−b)(y−b)=r2(x_1-a)(x-a)+(y_1-b)(y-b)=r^2，直接代入即可。

答题过程

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For the circle (x−2)2+(y−5)2=100(x - 2)^2 + (y - 5)^2 = 100, the tangent at P(x1,y1)P(x_1, y_1) is:

(x1−2)(x−2)+(y1−5)(y−5)=100(x_1 - 2)(x - 2) + (y_1 - 5)(y - 5) = 100

At P(8,−3)P(8, -3):

(8−2)(x−2)+(−3−5)(y−5)= 1006(x−2)+(−8)(y−5)= 1006x−12−8y+40= 1006x−8y+28= 1006x−8y−72= 03x−4y−36= 0\begin{align*} (8 - 2)(x - 2) + (-3 - 5)(y - 5) =&\, 100 \\ 6(x - 2) + (-8)(y - 5) =&\, 100 \\ 6x - 12 - 8y + 40 =&\, 100 \\ 6x - 8y + 28 =&\, 100 \\ 6x - 8y - 72 =&\, 0 \\ 3x - 4y - 36 =&\, 0 \end{align*} 3x−4y−36=0\boxed{3x - 4y - 36 = 0}