题目

Problem

A circle has equation $(x - 1)^2 + (y + 4)^2 = 40$ . A line with equation $y = x - 9$ intersects the circle at points $A$ and $B$ .

(a) Find the coordinates of the two points of intersection.

[4]

(b) Find an equation of the circle with diameter $AB$ .

[3]

题目中文翻译

一个圆的方程为 $(x - 1)^2 + (y + 4)^2 = 40$ 。方程为 $y = x - 9$ 的直线与该圆相交于点 $A$ 和 $B$ 。

(a) 求两个交点的坐标。

[4]

(b) 求以 $AB$ 为直径的圆的方程。

[3]

解答

展开

把直线的表达式代入圆方程。解出两个横坐标后，各自回代求纵坐标。

展开

Substituting $y=x-9$ gives $(x-1)^2+(x-5)^2=40$ , or

x^2-6x-7=(x+1)(x-7)=0.

For $x=-1$ , $y=-10$ . For $x=7$ , $y=-2$ . The intersections are $(-1,-10)$ and $(7,-2)$ .

展开

直径中点是新圆的圆心。从中点到任一端点的距离就是新半径。

展开

The midpoint of the two intersections is $(3,-6)$ . The squared radius is $(7-3)^2+(-2+6)^2=32$ . Hence the required circle is

(x-3)^2+(y+6)^2=32.