Course: 10.3 以原产地为中心圆环

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以起源居中圆圆
You draw a circle that is centered at the origin. You measure the diameter of the circle to be 32 units. Does the point $\begin{align*}(14, 8)\end{align*}$ lie on the circle?
::您绘制的圆以原点为中心。您测量圆的直径为 32 单位。点( 14, 8) 是否在圆上 ?

Circles Centered at the Origin
::以起源居中圆圆

Until now, your only reference to circles was from geometry. A circle is the set of points that are equidistant (the radius ) from a given point (the center ). A line segment that passes through the center and has endpoints on the circle is a diameter .
::直到现在,您对圆圈的唯一引用来自几何。一个圆圈是从给定点(中点)到给定点(中点)的等距(半径)的一组点。横穿中间并有圆圈端点的线段是直径。

Now, we will take a circle and place it on the $\begin{align*}x-y\end{align*}$ plane to see if we can find its equation. In this concept, we are going to place the center of the circle on the origin.
::现在,我们将进行一个圆圈,然后把它放在X-y平面上,看看我们能否找到它的方程。在这个概念中,我们将把圆圈的中心放在原点上。

Finding the Equation of a Circle
::查找圆形的等量

Step 1: On a piece of graph paper, draw an $\begin{align*}x-y\end{align*}$ plane. Using a compass, draw a circle, centered at the origin that has a radius of 5. Find the point $\begin{align*}(3, 4)\end{align*}$ on the circle and draw a right triangle with the radius as the hypotenuse.
::第1步:在一张图纸上,绘制一个 x -y 平面。使用一个罗盘,绘制一个圆形,以半径为 5 的原点为中心。在圆上找到点( 3, 4) , 并绘制一个以半径为下限的右三角形。

Step 2: Using the length of each side of the right triangle, show that the Pythagorean Theorem is true.
::第2步:使用右三角形每一侧的长度,显示毕达哥里安定理词是真实的。

Step 3: Now, instead of using $\begin{align*}(3, 4)\end{align*}$ , change the point to $\begin{align*}(x, y)\end{align*}$ so that it represents any point on the circle. Using $\begin{align*}r\end{align*}$ to represent the radius, rewrite the Pythagorean Theorem.
::步骤 3 : 现在, 不使用 (3, 4) , 将点更改为 (x,y) , 使其代表圆上的任何点。使用 r 代表半径, 重写 Pytagorean Theorem 。

The equation of a circle , centered at the origin, is $\begin{align*}x^2+y^2=r^2\end{align*}$ , where $\begin{align*}r\end{align*}$ is the radius and $\begin{align*}(x, y)\end{align*}$ is any point on the circle.
::圆的方程式,以原圆为中心,是 x2+y2=r2,其中 r 是半径, (x,y) 是圆上的任何点。

Let's find the radius of $\begin{align*}x^2+y^2=16\end{align*}$ and graph.
::让我们找到 x2+y2=16 和图形的半径。

To find the radius, we can set $\begin{align*}16=r^2\end{align*}$ , making $\begin{align*}r = 4\end{align*}$ . $\begin{align*}r\end{align*}$ is not -4 because it is a distance and distances are always positive. To graph the circle, start at the origin and go out 4 units in each direction and connect.
::要找到半径, 我们可以设置 16 = r2, make r= 4. r is not - 4, 因为它是一个距离, 距离总是正数。要绘制圆形图, 从起始点开始, 从每个方向取出 4 个单位并连接。

Now, let's find the equation of the circle with center at the origin and passes through $\begin{align*}(-7, -7)\end{align*}$ .
::现在,让我们找到圆形的方程式, 以中心为起点, 并通过(-7, -7) 。

Using the equation of the circle, we have: $\begin{align*}(-7)^2+(-7)^2=r^2\end{align*}$ . Solve for $\begin{align*}r^2\end{align*}$ .
::使用圆的方程式,我们有-7)2+(-7)2=r2. r2 解决 r2 。

$\begin{aligned} (- 7)^{2} + (- 7)^{2} & = r^{2} \\ 49 + 49 & = r^{2} \\ 98 & = r^{2} \end{aligned}$
$\begin{align*}(-7)^2+(-7)^2&=r^2 \\ 49+49&=r^2 \\ 98&=r^2\end{align*}$
:-7)2+(-7)2=r249+49=r298=r2

So, the equation is $\begin{align*}x^2+y^2=98\end{align*}$ . The radius of the circle is $\begin{align*}r=\sqrt{98}=7 \sqrt{2}\end{align*}$ .
::方程式是 x2+y2=98。圆的半径是 r=98=72。

Finally, let's determine if the point $\begin{align*}(9, -11)\end{align*}$ is on the circle $\begin{align*}x^2+y^2=225\end{align*}$ .
::最后,让我们确定点(9,-11)是否在x2+y2=225圆上。

Substitute the point in for $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ and see if it equals 225.
::替换 x 和 y 的点, 看看它是否等于 225 。

$\begin{aligned} 9^{2} + (- 11)^{2} & = 225 \\ 81 + 121 & \overset{?}{=} 225 \\ 202 & \neq 225 \end{aligned}$
$\begin{align*}9^2+(-11)^2&=225 \\ 81+121&\overset{?}{=}225 \\ 202& \ne 225\end{align*}$

The point is not on the circle.
::点不在圆圈上。

Examples
::实例

Example 1
::例1

Earlier, you were asked to determine if the point $\begin{align*}(14, 8)\end{align*}$ lies on the circle that is centered at the origin and has a diameter of 32 units.
::早些时候,有人要求你确定点(14,8)是否位于以原点为中心、直径为32个单位的圆上。

From this lesson, you know that the equation of a circle that is centered at the origin is $\begin{align*}x^2+y^2=r^2\end{align*}$ , where $\begin{align*}r\end{align*}$ is the radius and $\begin{align*}(x, y)\end{align*}$ is any point on the circle.
::从此课中, 您知道, 以源代码为中心的一个圆的方程式是 x2+y2=r2, 其中 r 是半径, (x,y) 是圆上的任何点。

With the point $\begin{align*}(14, 8)\end{align*}$ , $\begin{align*}x = 14\end{align*}$ and $\begin{align*}y = 8\end{align*}$ . We are given the diameter, but we need the radius. Recall that the radius is half the diameter, so the radius is $\begin{align*}\frac{32}{2} = 16\end{align*}$ .
::点数( 14, 8) x=14 和 Y=8 。我们得到了直径, 但我们需要半径。提醒注意半径是直径的一半, 所以半径是 322=16 。

Plug these values into the equation of the circle. If they result in a true statement, the point lies on the circle.
::将这些值插入圆的方程中。如果这些值产生真实的语句,则点在圆上。

$\begin{aligned} x^{2} + y^{2} = r^{2} \\ 14^{2} + 8^{2} \overset{?}{=} 16^{2} \\ 196 + 64 \overset{?}{=} 256 \\ 260 \neq 256 \end{aligned}$
$\begin{align*}x^2+y^2=r^2\\ 14^2 + 8^2 \overset{?}{=} 16^2\\ 196 + 64 \overset{?}{=}256\\ 260 \ne 256\end{align*}$
::x2+y2=r2142+82=? 162196+64=? 256260256

Therefore the point does not lie on the circle.
::因此,问题不在于圆圈上。

Example 2
::例2

Graph and find the radius of $\begin{align*}x^2+y^2=4\end{align*}$ .
::图形并查找 x2+y2=4 的半径。

$\begin{align*}r=\sqrt{4}=2\end{align*}$
::r=4=2

Example 3
::例3

Find the equation of the circle with a radius of $\begin{align*}6 \sqrt{5}\end{align*}$ .
::查找圆形的方程,半径为65。

Plug in $\begin{align*}6 \sqrt{5}\end{align*}$ for $\begin{align*}r\end{align*}$ in $\begin{align*}x^2+y^2=r^2\end{align*}$
::x2+y2=r2 中 r 的 r 在65 中插插件

$\begin{aligned} x^{2} + y^{2} & = {(6 \sqrt{5})}^{2} \\ x^{2} + y^{2} & = 6^{2} \cdot {(\sqrt{5})}^{2} \\ x^{2} + y^{2} & = 36 \cdot 5 \\ x^{2} + y^{2} & = 180 \end{aligned}$
$\begin{align*}x^2+y^2&=\left(6 \sqrt{5}\right)^2 \\ x^2+y^2&=6^2 \cdot \left(\sqrt{5}\right)^2 \\ x^2+y^2&=36 \cdot 5 \\ x^2+y^2&=180\end{align*}$
::x2+y2 = (652)2x2+y2=62*(5)2x2+y2=365x2+y2=180

Example 4
::例4

Find the equation of the circle that passes through $\begin{align*}(5, 8)\end{align*}$ .
::查找通过(5,8)的圆形的方程。

Plug in $\begin{align*}(5, 8)\end{align*}$ for $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ , respectively.
::x和y分别插在(5,8)内。

$\begin{aligned} 5^{2} + 8^{2} & = r^{2} \\ 25 + 64 & = r^{2} \\ 89 & = r^{2} \end{aligned}$
$\begin{align*}5^2+8^2&=r^2 \\ 25+64&=r^2 \\ 89&=r^2\end{align*}$
::52+82=r225+64=r289=r2

The equation is $\begin{align*}x^2+y^2=89\end{align*}$
::方程式是 x2+y2=89

Example 5
::例5

Determine if $\begin{align*}(-10, 7)\end{align*}$ is on the circle $\begin{align*}x^2+y^2=149\end{align*}$ .
::确定是否 (- 10, 7) 在 x2+y2=149 圆上。

Plug in $\begin{align*}(-10, 7)\end{align*}$ to see if it is a valid equation.
::插在(- 10, 7) 中, 看看它是否是一个有效的方程。

$\begin{aligned} (- 10)^{2} + 7^{2} & = 149 \\ 100 + 49 & = 149 \end{aligned}$
$\begin{align*}(-10)^2+7^2&=149 \\ 100+49&=149\end{align*}$

Yes, the point is on the circle.
::是的,重点是在圆圈上。

Review
::回顾

Graph the following circles and find the radius.
::绘制以下圆形图并找到半径。

$\begin{align*}x^2+y^2=9\end{align*}$
::x2+y2=9

$\begin{align*}x^2+y^2=64\end{align*}$
::x2+y2=64

$\begin{align*}x^2+y^2=8\end{align*}$
::x2+y2=8

$\begin{align*}x^2+y^2=50\end{align*}$
::x2+y2=50

$\begin{align*}2x^2+2y^2=162\end{align*}$
::2x2+2y2=162

$\begin{align*}5x^2+5y^2=150\end{align*}$
::5x2+5y2=150

Write the equation of the circle with the given radius and centered at the origin.
::以给定半径写出圆形的方程式, 以原点为中心。

14

6

$\begin{align*}9 \sqrt{2}\end{align*}$

Write the equation of the circle that passes through the given point and is centered at the origin.
::写入圆的方程式,该方程式通过给定点,以原点为中心。

$\begin{align*}(7,-24)\end{align*}$

$\begin{align*}(2,2)\end{align*}$

$\begin{align*}(-9,-10)\end{align*}$

Determine if the following points are on the circle, $\begin{align*}x^2+y^2=74\end{align*}$ .
::确定以下点是否在圆上, x2+y2=74。

$\begin{align*}(-8,0)\end{align*}$

$\begin{align*}(7,-5)\end{align*}$

$\begin{align*}(6,-6)\end{align*}$

Challenge In Geometry, you learned about tangent lines to a circle. Recall that the tangent line touches a circle at one point and is perpendicular to the radius at that point, called the point of tangency.
::在几何学的挑战中, 您学习到一个圆形的正切线。回顾正切线在一个点上碰到一个圆形, 并且与那个点的半径垂直, 叫做切视点。

The equation of a circle is $\begin{align*}x^2+y^2=10\end{align*}$ with point of tangency $\begin{align*}(-3, 1)\end{align*}$ .

Find the slope of the radius from the center to $\begin{align*}(-3, 1)\end{align*}$ .
::查找半径的斜坡,从中间到(- 3, 1) 。

Find the perpendicular slope to (a). This is the slope of the tangent line.
::找到(a)的垂直斜坡。这是正切线的斜坡。

Use the slope from (b) and the given point to find the equation of the tangent line.
::使用 (b) 的斜坡和给定点查找正切线的方程。

::圆的方程式是x2+y2=10,并带有相干点(- 3, 1) 。查找半径的斜度, 从中间到 (- 3, 1) 。查找垂直斜度到 (a) 。这是正切线的斜度。使用从 (b) 和给定点的斜度来查找正切线的方形。

Repeat the steps in #16 to find the equation of the tangent line to $\begin{align*}x^2+y^2=34\end{align*}$ with a point of tangency of $\begin{align*}(3, 5)\end{align*}$ .
::重复 # 16 中的步骤, 以找到相切线的方程为 x2+y2=34, 相切点为( 3 5) 。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

以起源居中圆圆

Circles Centered at the Origin ::以起源居中圆圆

Finding the Equation of a Circle ::查找圆形的等量

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Circles Centered at the Origin
::以起源居中圆圆

Finding the Equation of a Circle
::查找圆形的等量

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)