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P2 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)
题目中文翻译

圆 CC 的方程为

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

(a) 求 (i) CC 的圆心坐标；(ii) CC 的半径的精确值。(3)

直线 LL 的方程为 y=mx+1y=mx+1，其中 mm 为常数。已知 LL 是 CC 在点 PP 处的切线。

(b) 证明：(3)

2m2−7m−22=02m^2-7m-22=0

(c) 由此求 PP 的所有可能的坐标对。(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=0 x2+m2x2+2mx+1+4x −30mx−30+209=0(1+m2)x2+(4−28m)x+180= 0.\begin{align*} &\,x^2+(mx+1)^2+4x\\ &\,\hspace{2pt}-30(mx+1)+209=0\\ &\,x^2+m^2x^2+2mx+1+4x\\ &\,\hspace{2pt}-30mx-30+209=0\\ (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\\ 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}\\ \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\\ (2m+14)^2 =&\, 20(m^2+1)\\ 4m^2 + 56m + 196 =&\, 20m^2 + 20\\ 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*}

Using the quadratic from part (b),

1254x2−150x+180=0.\frac{125}{4}x^2-150x+180=0.

Multiply by 4/54/5 and factorise:

25x2−120x+144= 0(5x−12)2= 0x= 125.\begin{align*} 25x^2-120x+144=&\,0\\ (5x-12)^2=&\,0\\ x=&\,\frac{12}{5}. \end{align*}

Then

y= 112(125)+1= 665+1= 715.\begin{align*} y=&\,\frac{11}{2}\bigg(\frac{12}{5}\bigg)+1\\ =&\,\frac{66}{5}+1\\ =&\,\frac{71}{5}. \end{align*}

The tangent point is (125,715)\big(\dfrac{12}{5},\dfrac{71}{5}\big).

If m=−2m=-2, then

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

Using the quadratic from part (b),

5x2+60x+180= 0(x+6)2= 0x= −6.\begin{align*} 5x^2+60x+180=&\,0\\ (x+6)^2=&\,0\\ x=&\,-6. \end{align*}

Then y=−2(−6)+1=13y=-2(-6)+1=13. The tangent point is (−6,13)(-6,13).

Therefore, the possible coordinates of PP are

(125,715)and(−6,13).\big(\frac{12}{5},\frac{71}{5}\big) \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)\\ 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 +16111121x+22= −4x+322125x= 300x= 125\begin{align*} &\,\frac{11}{2}x+1\\ =&\,-\frac{2}{11}x\\ &\,\hspace{2pt}+\frac{161}{11}\\ 121x+22=&\,-4x+322\\ 125x=&\,300\\ x =&\, \frac{12}{5} \end{align*}

Substituting x=125x=\dfrac{12}{5} into the tangent gives y=715y=\dfrac{71}{5}. Thus, one possible point is P(125,715)P\big(\frac{12}{5}, \frac{71}{5}\big).

  • 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)\\ 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\\ -\frac{5}{2}x =&\, 15\\ 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*} \big(\frac{12}{5},\frac{71}{5}\big) \quad\text{and}\quad (-6,13) \end{align*}

解法三

思路

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两条切线都经过 T(0,1)T(0,1)。半径垂直于切线，因而圆心、切点和 TT 构成直角三角形；先用勾股求切线段长度，再求坐标，并把正负候选代回圆方程筛选。

答题过程

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From part (b), m=112m=\dfrac{11}{2} or m=−2m=-2. Both tangents pass through T(0,1)T(0,1). Let the centre be A(−2,15)A(-2,15).

AT2=(−2)2+(15−1)2=200.AT^2=(-2)^2+(15-1)^2=200.

Since APAP is perpendicular to TPTP and AP2=20AP^2=20,

TP2=AT2−AP2=180.TP^2=AT^2-AP^2=180.

For m=−2m=-2, write P=(a,1−2a)P=(a,1-2a). Then

TP2= a2+(−2a)25a2= 180a= ±6.\begin{align*} TP^2=&\,a^2+(-2a)^2\\ 5a^2=&\,180\\ a=&\,\pm6. \end{align*}

The candidate (6,−11)(6,-11) is not on the circle:

(6+2)2+(−11−15)2=740≠20.(6+2)^2+(-11-15)^2=740\neq20.

The candidate (−6,13)(-6,13) satisfies (x+2)2+(y−15)2=20(x+2)^2+(y-15)^2=20, so it is the tangent point.

For m=112m=\dfrac{11}{2}, write P=(b,1+112b)P=\big(b,1+\dfrac{11}{2}b\big). Then

TP2= b2+(112b)21254b2= 180b2= 14425b= ±125.\begin{align*} TP^2=&\,b^2+\bigg(\frac{11}{2}b\bigg)^2\\ \frac{125}{4}b^2=&\,180\\ b^2=&\,\frac{144}{25}\\ b=&\,\pm\frac{12}{5}. \end{align*}

The negative candidate (−125,−615)\big(-\dfrac{12}{5},-\dfrac{61}{5}\big) is not on the circle:

 (−125+2)2 +(−615−15)2= 740≠20.\begin{align*} &\,\bigg(-\frac{12}{5}+2\bigg)^2\\ &\,\hspace{2pt}+\bigg(-\frac{61}{5}-15\bigg)^2\\ =&\,740\neq20. \end{align*}

The positive candidate (125,715)\big(\dfrac{12}{5},\dfrac{71}{5}\big) lies on the circle:

 (125+2)2 +(715−15)2= 20.\begin{align*} &\,\bigg(\frac{12}{5}+2\bigg)^2\\ &\,\hspace{2pt}+\bigg(\frac{71}{5}-15\bigg)^2\\ =&\,20. \end{align*}

Thus the two possible points are

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