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IAL 2024 May Q7

A Level / Edexcel / P2

IAL 2024 May Paper · Question 7

题目

Problem

The circle C1C_1 has equation

x2+y2+8x−10y=29x^2+y^2+8x-10y=29

(a) (i) Find the coordinates of the centre of C1C_1

(ii) Find the exact value of the radius of C1C_1

(3)

In part (b) you must show detailed reasoning.

The circle C2C_2 has equation

(x−5)2+(y+8)2=52(x-5)^2+(y+8)^2=52

(b) Prove that the circles C1C_1 and C2C_2 neither touch nor intersect.

(3)

解答

(a)

解法一

思路

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把 x2+8xx^2+8x 和 y2−10yy^2-10y 分别配方，写成圆的标准形式。

答题过程

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Complete the square:

x2+y2+8x−10y= 29(x+4)2−16+(y−5)2−25= 29(x+4)2+(y−5)2= 70.\begin{align*} x^2+y^2+8x-10y=&\,29\\ (x+4)^2-16+(y-5)^2-25=&\,29\\ (x+4)^2+(y-5)^2=&\,70. \end{align*}

Therefore, the centre of C1C_1 is

(−4,5)\begin{align*} (-4,5) \end{align*}

and the radius is

70.\begin{align*} \sqrt{70}. \end{align*}

(b)

解法一

思路

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两个圆若相交或相切，圆心距必须不大于两半径之和。这里比较圆心距和半径之和即可。

答题过程

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For C1C_1,

centre=(−4,5),r1=70.\begin{align*} \text{centre}=(-4,5),\qquad r_1=\sqrt{70}. \end{align*}

For C2C_2,

centre=(5,−8),r2=52.\begin{align*} \text{centre}=(5,-8),\qquad r_2=\sqrt{52}. \end{align*}

The distance between the centres is

d= (5−(−4))2+(−8−5)2= 92+(−13)2= 250= 15.81….\begin{align*} d=&\,\sqrt{(5-(-4))^2+(-8-5)^2}\\[4mm] =&\,\sqrt{9^2+(-13)^2}\\[4mm] =&\,\sqrt{250}\\[4mm] =&\,15.81\ldots. \end{align*}

The sum of the radii is

r1+r2= 70+52= 15.58….\begin{align*} r_1+r_2=&\,\sqrt{70}+\sqrt{52}\\[4mm] =&\,15.58\ldots. \end{align*}

Since

250>70+52,\begin{align*} \sqrt{250}>\sqrt{70}+\sqrt{52}, \end{align*}

the distance between the centres is greater than the sum of the radii.

Therefore, the circles neither touch nor intersect.

解法二

思路

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代数判别式法。若两圆相交或相切，联立两圆的方程应有实数解。两圆方程相减可以消去二次项，得到一条直线方程（根轴）。将此直线代入任意圆的方程，得到关于一个变量的一元二次方程。通过计算其判别式 Δ<0\Delta < 0 证明方程无实数解，即可说明两圆既不相切也不相交。

答题过程

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The equations of the two circles are:

C1:  x2+y2+8x−10y−29=0C2:  (x−5)2+(y+8)2=52  ⟹  x2+y2−10x+16y+37=0\begin{align*} C_1:&\,\, x^2+y^2+8x-10y-29 = 0\\[4mm] C_2:&\,\, (x-5)^2+(y+8)^2 = 52 \implies x^2+y^2-10x+16y+37 = 0 \end{align*}

Subtract C2C_2 from C1C_1 to eliminate the x2x^2 and y2y^2 terms:

(8x−10y−29)−(−10x+16y+37)=  018x−26y−66=  09x−13y−33=  0  ⟹  x=13y+339\begin{align*} (8x-10y-29) - (-10x+16y+37) =&\,\, 0\\[4mm] 18x - 26y - 66 =&\,\, 0\\[4mm] 9x - 13y - 33 =&\,\, 0 \implies x = \frac{13y+33}{9} \end{align*}

Substitute x=13y+339x = \frac{13y+33}{9} into the equation of C1C_1:

(13y+339)2+y2+8(13y+339)−10y−29=0\begin{align*} \left(\frac{13y+33}{9}\right)^2 + y^2 + 8\left(\frac{13y+33}{9}\right) - 10y - 29 = 0 \end{align*}

Multiply the entire equation by 8181 to clear the denominator:

(13y+33)2+81y2+72(13y+33)−810y−2349=0\begin{align*} (13y+33)^2 + 81y^2 + 72(13y+33) - 810y - 2349 = 0 \end{align*}

Expand and simplify:

(169y2+858y+1089)+81y2+(936y+2376)−810y−2349=  0250y2+984y+1116=  0125y2+492y+558=  0\begin{align*} (169y^2 + 858y + 1089) + 81y^2 + (936y + 2376) - 810y - 2349 =&\,\, 0\\[4mm] 250y^2 + 984y + 1116 =&\,\, 0\\[4mm] 125y^2 + 492y + 558 =&\,\, 0 \end{align*}

Calculate the discriminant Δ\Delta of this quadratic equation:

Δ=  b2−4ac=  4922−4(125)(558)=  242064−279000=  −36936\begin{align*} \Delta =&\,\, b^2 - 4ac\\[4mm] =&\,\, 492^2 - 4(125)(558)\\[4mm] =&\,\, 242064 - 279000\\[4mm] =&\,\, -36936 \end{align*}

Since the discriminant Δ<0\Delta < 0, the quadratic equation has no real solutions.

This means there are no real points of intersection between the two circles.

Therefore, the circles neither touch nor intersect.