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IAL 2024 Oct Q10

A Level / Edexcel / P2

IAL 2024 Oct Paper · Question 10

题目

Problem

The circle CC has equation

x2+y2+4x−30y+209=0x^2+y^2+4x-30y+209=0

(a) Find

(i) the coordinates of the centre of CC,

(ii) the exact value of the radius of CC.

(3)

The line LL has equation y=mx+1y=mx+1, where mm is a constant. Given that LL is the tangent to CC at the point PP,

(b) show that

2m2−7m−22=02m^2-7m-22=0
(3)

(c) Hence find the possible pairs of coordinates of PP.

(4)

解答

(a)

解法一

思路

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配方成圆的标准形式 (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2。

答题过程

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Complete the square:

x2+4x+y2−30y+209= 0(x+2)2−4+(y−15)2−225+209= 0(x+2)2+(y−15)2= 20.\begin{align*} x^2+4x+y^2-30y+209=&\,0\\ (x+2)^2-4+(y-15)^2-225+209=&\,0\\ (x+2)^2+(y-15)^2=&\,20. \end{align*}

Therefore, the centre is

(−2,15)\begin{align*} (-2,15) \end{align*}

and the radius is

20=25.\begin{align*} \sqrt{20}=2\sqrt5. \end{align*}

(b)

解法一

思路

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切线与圆只有一个交点，所以把 y=mx+1y=mx+1 代入圆方程后，所得关于 xx 的二次方程判别式为 00。

答题过程

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Substitute y=mx+1y=mx+1 into the circle:

x2+(mx+1)2+4x−30(mx+1)+209=  0x2+m2x2+2mx+1+4x−30mx−30+209=  0(1+m2)x2+(4−28m)x+180=  0.\begin{align*} x^2+(mx+1)^2+4x-30(mx+1)+209 =&\,\, 0\\[4mm] x^2+m^2x^2+2mx+1+4x-30mx-30+209 =&\,\, 0\\[4mm] (1+m^2)x^2+(4-28m)x+180 =&\,\, 0. \end{align*}

Since the line is tangent to the circle, this quadratic has equal roots. Therefore,

(4−28m)2−4(1+m2)(180)=  0.\begin{align*} (4-28m)^2-4(1+m^2)(180) = \,\, 0. \end{align*}

Expand:

16−224m+784m2−720−720m2=  064m2−224m−704=  0.\begin{align*} 16-224m+784m^2-720-720m^2 =&\,\, 0\\[4mm] 64m^2-224m-704 =&\,\, 0. \end{align*}

Divide by 3232:

2m2−7m−22=0.\begin{align*} 2m^2-7m-22=0. \end{align*}

解法二

思路

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点到直线距离公式。设切线方程为 mx−y+1=0mx-y+1=0。圆的切线到圆心的垂直距离等于圆的半径。利用点到直线距离公式 d=∣ax0+by0+c∣a2+b2d = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}，将圆心 (−2,15)(-2, 15) 和半径 20\sqrt{20} 代入建立方程，平方后即可直接求出关于 mm 的二次方程。

答题过程

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The equation of the line LL is:

y=mx+1  ⟹  mx−y+1=0\begin{align*} y = mx+1 \implies mx-y+1 = 0 \end{align*}

From part (a), the circle has centre A(−2,15)A(-2, 15) and radius r=20r = \sqrt{20}.

Since LL is a tangent to the circle, the perpendicular distance from the centre AA to the line LL must equal the radius rr:

∣m(−2)−(15)+1∣m2+(−1)2=20\begin{align*} \frac{|m(-2) - (15) + 1|}{\sqrt{m^2 + (-1)^2}} = \sqrt{20} \end{align*}

Simplify the numerator:

∣−2m−14∣m2+1=  20∣2m+14∣m2+1=  20\begin{align*} \frac{|-2m - 14|}{\sqrt{m^2 + 1}} =&\,\, \sqrt{20}\\[4mm] \frac{|2m + 14|}{\sqrt{m^2 + 1}} =&\,\, \sqrt{20} \end{align*}

Square both sides:

(2m+14)2m2+1=  20(2m+14)2=  20(m2+1)4m2+56m+196=  20m2+2016m2−56m−176=  0\begin{align*} \frac{(2m + 14)^2}{m^2 + 1} =&\,\, 20\\[4mm] (2m+14)^2 =&\,\, 20(m^2+1)\\[4mm] 4m^2 + 56m + 196 =&\,\, 20m^2 + 20\\[4mm] 16m^2 - 56m - 176 =&\,\, 0 \end{align*}

Divide the entire equation by 88:

2m2−7m−22=0\begin{align*} 2m^2 - 7m - 22 = 0 \end{align*}

This is the required result.

(c)

解法一

思路

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先解出两个可能的切线斜率，再分别代入圆与直线求切点坐标。

答题过程

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Solve

2m2−7m−22= 0(2m−11)(m+2)= 0.\begin{align*} 2m^2-7m-22=&\,0\\ (2m-11)(m+2)=&\,0. \end{align*}

So

m=112orm=−2.\begin{align*} m=\frac{11}{2}\quad\text{or}\quad m=-2. \end{align*}

If m=112m=\frac{11}{2}, then

y=112x+1.\begin{align*} y=\frac{11}{2}x+1. \end{align*}

Substitution gives the tangent point

(125,715).\begin{align*} \left(\frac{12}{5},\frac{71}{5}\right). \end{align*}

If m=−2m=-2, then

y=−2x+1.\begin{align*} y=-2x+1. \end{align*}

Substitution gives the tangent point

(−6,13).\begin{align*} (-6,13). \end{align*}

Therefore, the possible coordinates of PP are

(125,715)and(−6,13).\left(\frac{12}{5},\frac{71}{5}\right) \quad\text{and}\quad (-6,13).

解法二

思路

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垂直法线交点求切点。已知切点 PP 处切线与半径垂直。过圆心 (−2,15)(-2,15) 且与切线垂直的直线（法线）必通过切点 PP。因此求出法线斜率 −1m-\frac{1}{m} 对应的方程后，分别与切线联立即可快速得到切点。

答题过程

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Solve the equation from part (b):

2m2−7m−22=0  ⟹  (2m−11)(m+2)=0\begin{align*} 2m^2-7m-22 = 0 \implies (2m-11)(m+2) = 0 \end{align*}

So the possible gradients of the tangents are:

m=112orm=−2\begin{align*} m = \frac{11}{2} \quad\text{or}\quad m = -2 \end{align*}

The normal to the circle at the tangent point PP must pass through the centre A(−2,15)A(-2, 15) and have a gradient of −1m-\frac{1}{m}.

  • Case 1: m=112m = \frac{11}{2} The gradient of the normal is −211-\frac{2}{11}. The equation of the normal is:
y−15=  −211(x+2)y=  −211x+16111\begin{align*} y - 15 =&\,\, -\frac{2}{11}(x+2)\\[4mm] y =&\,\, -\frac{2}{11}x + \frac{161}{11} \end{align*}

Equate the tangent y=112x+1y = \frac{11}{2}x + 1 and the normal:

112x+1=  −211x+16111(112+211)x=  16111−112522x=  15011x=  125\begin{align*} \frac{11}{2}x + 1 =&\,\, -\frac{2}{11}x + \frac{161}{11}\\[4mm] \left( \frac{11}{2} + \frac{2}{11} \right)x =&\,\, \frac{161}{11} - 1\\[4mm] \frac{125}{22}x =&\,\, \frac{150}{11}\\[4mm] x =&\,\, \frac{12}{5} \end{align*}

Substitute x=125x = \frac{12}{5} into the tangent equation: y=112(125)+1=715y = \frac{11}{2}\left(\frac{12}{5}\right) + 1 = \frac{71}{5}. Thus, one possible point is P(125,715)P\left(\frac{12}{5}, \frac{71}{5}\right).

  • Case 2: m=−2m = -2 The gradient of the normal is 12\frac{1}{2}. The equation of the normal is:
y−15=  12(x+2)y=  12x+16\begin{align*} y - 15 =&\,\, \frac{1}{2}(x+2)\\[4mm] y =&\,\, \frac{1}{2}x + 16 \end{align*}

Equate the tangent y=−2x+1y = -2x + 1 and the normal:

−2x+1=  12x+16−52x=  15x=  −6\begin{align*} -2x + 1 =&\,\, \frac{1}{2}x + 16\\[4mm] -\frac{5}{2}x =&\,\, 15\\[4mm] x =&\,\, -6 \end{align*}

Substitute x=−6x = -6 into the tangent equation: y=−2(−6)+1=13y = -2(-6) + 1 = 13. Thus, the other possible point is P(−6,13)P(-6, 13).

Therefore, the possible pairs of coordinates of PP are:

(125,715)and(−6,13)\begin{align*} \left(\frac{12}{5},\frac{71}{5}\right) \quad\text{and}\quad (-6,13) \end{align*}