题目

Problem

A circle has equation

x^2+y^2+8x-14y-79=0

(a) Find

(i) the coordinates of the centre of the circle,

(ii) the radius of the circle.

(3)

Given that $P$ is the point on the circle that is nearest the origin $O$ ,

(b) find the exact length of $OP$

(2)

解答

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对 $x$ 和 $y$ 分别配方，化成 $(x-a)^2+(y-b)^2=r^2$ 的形式，直接读出圆心和半径。

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Completing the square for both $x$ and $y$ ,

\begin{align*} x^2+8x+y^2-14y=&\,79\\[2mm] (x+4)^2-16+(y-7)^2-49=&\,79\\[2mm] (x+4)^2+(y-7)^2=&\,144. \end{align*}

(i) The centre is $(-4,\,7)$ .

(ii) The radius is $\sqrt{144}=12$ .

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原点 $O$ 到圆心 $C$ 的距离是 $OC$ 。 $P$ 是圆上离 $O$ 最近的点，所以 $O$ 、 $P$ 、 $C$ 三点共线， $OP=|OC-r|$ 。

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The centre is $C(-4,7)$ and the radius is $r=12$ .

The distance from the origin $O$ to $C$ is

OC=\sqrt{(-4)^2+7^2}=\sqrt{16+49}=\sqrt{65}.

Since $\sqrt{65}<12$ , the origin lies inside the circle. The point $P$ on the circle nearest to $O$ lies on the line $OC$ , so

OP=r-OC=12-\sqrt{65}.