Section: 居于 (h, k) 的圆圆 | 10.4 圆环居于(h, k)

Section outline

You draw a circle that is centered at $\begin{align*}(-2, -2)\end{align*}$ . You measure the diameter of the circle to be 18 units. Does the point $\begin{align*}(4, 5)\end{align*}$ lie on the circle?
::您绘制了一个以 (- 2) 为中心的圆形。您测量圆形的直径为 18 个单位。点( 4 5) 是否在圆形上 ?

Circles Centered at (h,k)
::居于 (h,k) 的圆圆

When a circle is centered at the origin, the equation is $\begin{align*}x^2+y^2=r^2\end{align*}$ . If we rewrite this equation, using the center , it would look like $\begin{align*}(x-0)^2+(y-0)^2=r^2\end{align*}$ . Extending this idea to any point as the center, we would have $\begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}$ , where $\begin{align*}(h, k)\end{align*}$ is the center.
::当圆以原点为中心时,方程式为 x2+y2=r2。如果我们使用中点重写此方程式,它看起来像(x-0)2+(y-0)2=r2) 。把这个想法扩展至中点的任何点,我们就会有(x-h)2+(y-k)2=r2, 其中(h,k)是中点。

Let's find the center and radius of $\begin{align*}(x+1)^2+(y-3)^2=16\end{align*}$ and graph.
::让我们找到( x+1) 2+(y-3) 2=16 和图形的中心和半径。

Using the general equation above, the center would be $\begin{align*}(-1, 3)\end{align*}$ and the radius is $\begin{align*}\sqrt{16}\end{align*}$ or 4. To graph, plot the center and then go out 4 units up, down, to the left, and to the right.
::使用上面的一般方程,中心将是 (-1,3) , 半径是 16 或 4 。要绘制图, 绘制中心, 然后从4个单元上、下、左、右, 向右。

Now, let's find the equation of the circle with center $\begin{align*}(2, 4)\end{align*}$ and radius 5.
::现在,让我们找到圆的方程,以中间(2,4)和半径5为中心。

Plug in the center and radius to the equation and simplify.
::插在方程的中间和半径,并简化。

$\begin{aligned} (x - 2)^{2} + (y - 4)^{2} & = 5^{2} \\ (x - 2)^{2} + (y - 4)^{2} & = 25 \end{aligned}$
$\begin{align*}(x-2)^2+(y-4)^2&=5^2 \\ (x-2)^2+(y-4)^2&=25\end{align*}$
:x-2)2+(y-4)2=52(x-2)2+(y-4)2=25

Finally, let's find the equation of the circle with center $\begin{align*}(6, -1)\end{align*}$ and $\begin{align*}(5, 2)\end{align*}$ is on the circle.
::最后,让我们找到圆的方程式, 以圆为中心( 6, - 1) 和( 5, 2) 在圆上。

In this problem , we are not given the radius. To find the radius, we must use the distance formula, $\begin{align*}d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\end{align*}$ .
::在此问题上, 我们没有得到半径。要找到半径, 我们必须使用距离公式 d= (x2- x1) 2+ (y2-y1) 2 。

$\begin{aligned} r & = \sqrt{{(5 - 6)}^{2} + {(2 - (- 1))}^{2}} \\ = \sqrt{{(- 1)}^{2} + 3^{2}} \\ = \sqrt{1 + 9} \\ = \sqrt{10} \end{aligned}$
$\begin{align*}r&=\sqrt{\left(5-6\right)^2+ \left(2- \left(-1\right)\right)^2} \\ &=\sqrt{\left(-1\right)^2+3^2} \\ &=\sqrt{1+9} \\ &=\sqrt{10}\end{align*}$
::r=( 5-6) 2+( 2- (-1) 1) 2=( 1) 2=( 1) 2+32=1+9=10

Therefore, the equation of this circle is $\begin{align*}(x-6)^2+(y-(-1))^2=\left(\sqrt{10}\right)^2\end{align*}$ or $\begin{align*}(x-6)^2+(y+1)^2=10\end{align*}$ .
::因此,这个圆的方程式是(x-6)2+(y-(-1))2=(10)2或(x-6)2+(y+1)2=10。

Examples
::实例

Example 1
::例1

Earlier, you were asked to determine if the point $\begin{align*}(4, 5)\end{align*}$ lies on the circle.
::早些时候,你被要求确定点(4,5)是否在圆上。

In this lesson, you learned the equation of a circle that is centered somewhere other than the origin is $\begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}$ , where $\begin{align*}(h, k)\end{align*}$ is the center.
::在此课中, 您会学到一个圆的方程式, 圆的中心点不是原点( x- h) 2+(y- k) 2=r2, 中心点是 (h, k) 。

We are given that the center is $\begin{align*}(-2, -2)\end{align*}$ , so $\begin{align*}h= -2\end{align*}$ and $\begin{align*}k = -2\end{align*}$ . We are also given the diameter of the circle, but we need the radius. Recall that the radius is half the diameter, so $\begin{align*}r = \frac{18}{2} = 9\end{align*}$ .
::我们被告知中心是 (-2,-2), 所以 h2 和 k2 。我们也被给出圆的直径, 但我们需要半径。回顾半径是直径的一半, 所以 r= 182=9 。

If we plug these values into the equation for the circle, we get:
::如果我们将这些值插入圆的方程,我们就会得到:

$\begin{aligned} (x - h)^{2} + (y - k)^{2} = r^{2} \\ (x - (- 2))^{2} + (y - (- 2))^{2} = 9^{2} \\ (x + 2)^{2} + (y + 2)^{2} = 81 \end{aligned}$
$\begin{align*}(x-h)^2+(y-k)^2=r^2\\ (x-(-2))^2 + (y - (-2))^2 = 9^2\\ (x + 2)^2 + (y + 2)^2 = 81\end{align*}$
:x-h)2+(y-k)2=r2(x-(-2))2+(y-2)2+(y-2)2=92(x+2)2+(y+2)2+(y+2)2=81

Now to find if the point $\begin{align*}(4, 5)\end{align*}$ lies on the circle we substitute 4 for x and 5 for y and see if the equation holds true.
::现在找到点( 4, 5) 是否位于我们替换 x 4 和 y 5 的圆上, 看看方程是否正确。

$\begin{aligned} (x + 2)^{2} + (y + 2)^{2} = 81 \\ (4 + 2)^{2} + (5 + 2)^{2} \overset{?}{=} 81 \\ 6^{2} + 7^{2} \overset{?}{=} 81 \\ 85 \neq 81 \end{aligned}$
$\begin{align*}(x + 2)^2 + (y + 2)^2 = 81\\ (4 + 2)^2 + (5 + 2)^2 \overset{?}{=} 81\\ 6^2 + 7^2 \overset{?}{=} 81\\ 85 \ne 81\end{align*}$
:x+2)2+2+(y+2)2=81(4+2)2+(5+2)2+(5+2)2=?8162+72=?8188581

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

Example 2
::例2

Graph $\begin{align*}(x+4)^2+(y+3)^2=25\end{align*}$ and find the center and radius.
::图表 (x+4) 2+(y+3) 2=25, 并找到中心与半径。

The center is $\begin{align*}(-4, -3)\end{align*}$ and the radius is 5.
::中心是(-4,-3),半径是5。

Example 3
::例3

Find the equation of the circle with center $\begin{align*}(-8, 3)\end{align*}$ and $\begin{align*}(2, -5)\end{align*}$ is on the circle.
::查找圆的方程式,以圆为中心(-8,3)和(-2,5)在圆上。

Use the distance formula to find the radius.
::使用距离公式找到半径。

$\begin{aligned} r & = \sqrt{{(2 - (- 8))}^{2} + {(- 5 - 3)}^{2}} \\ = \sqrt{10^{2} + {(- 8)}^{2}} \\ = \sqrt{100 + 64} \\ = \sqrt{164} \end{aligned}$
$\begin{align*}r&=\sqrt{\left(2- \left(-8\right)\right)^2+ \left(-5-3\right)^2}\\ &=\sqrt{10^2+ \left(-8\right)^2} \\ &=\sqrt{100+64} \\ &=\sqrt{164}\end{align*}$
::r=(2-(-8))2+(-5-3)2=102+(-8)2=100+64=164

The equation of this circle is $\begin{align*}(x+8)^2+(y-3)^2=164\end{align*}$ .
::此圆的方程式是 (x+8) 2+(y-3)2=164。

Example 4
::例4

The endpoints of a diameter of a circle are $\begin{align*}(-3, 1)\end{align*}$ and $\begin{align*}(9, 6)\end{align*}$ . Find the equation.
::圆形直径的终点是(-3,1)和(9,6),找出方程。

In this example , we are not given the center or radius. We can find the length of the diameter using the distance formula and then divide it by 2.
::在此示例中, 我们没有得到中心或半径。我们可以使用距离公式找到直径的长度, 然后将其除以 2 。

$\begin{aligned} d & = \sqrt{{(9 - (- 3))}^{2} + {(6 - 1)}^{2}} \\ = \sqrt{12^{2} + 5^{2}} The radius is 13 \div 2 = \frac{13}{2} . \\ = \sqrt{144 + 25} \\ = \sqrt{169} = 13 \end{aligned}$
$\begin{align*}d&=\sqrt{\left(9- \left(-3\right)\right)^2+ \left(6-1\right)^2} \\ &=\sqrt{12^2+5^2} \qquad \text{The radius is} \ 13 \div 2=\frac{13}{2}. \\ &=\sqrt{144+25} \\ &=\sqrt{169}=13\end{align*}$
::d=( 9- (- 3)) 2+( 6- 1) 2=122+52 半径为 132=132. =144+25=169=13

Now, we need to find the center. Use the midpoint formula with the endpoints.
::现在,我们需要找到中间点。使用中点公式和端点。

$\begin{aligned} c & = (\frac{- 3 + 9}{2}, \frac{1 + 6}{2}) \\ = (3, \frac{7}{2}) \end{aligned}$
$\begin{align*}c&=\left(\frac{-3+9}{2}, \frac{1+6}{2}\right) \\ &=\left(3, \frac{7}{2}\right)\end{align*}$
::c=(-3+92,1+62)=(3,72)

Therefore, the equation is $\begin{align*}(x-3)^2+ \left(y- \frac{7}{2}\right)^2=\frac{169}{4}\end{align*}$ .
::因此,等式是(x-3-3)2+(y-72)2=1694。

Review
::回顾

For questions 1-4, match the equation with the graph.
::对于问题14, 将方程式与图表匹配。

$\begin{align*}(x-8)^2+(y+2)^2=4\end{align*}$
:x-8)2+(y+2)2=4

$\begin{align*}x^2+(y-6)^2=9\end{align*}$
::x2+(y-6)2=9

$\begin{align*}(x+2)^2+(y-3)^2=36\end{align*}$
:x+2)2+(y-3)2=36

$\begin{align*}(x-4)^2+(y+4)^2=25\end{align*}$
:x-4)2+(y+4)2=25

Graph the following circles. Find the center and radius.
::绘制以下圆形图,找到中间和半径

$\begin{align*}(x-2)^2+(y-5)^2=16\end{align*}$
:x-2)2+(y-5)2=16

$\begin{align*}(x+4)^2+(y+3)^2=18\end{align*}$
:x+4)2+(y+3)2=18

$\begin{align*}(x+7)^2+(y-1)^2=8\end{align*}$
:x+7)2+(y-1)2=8

Find the equation of the circle, given the information below.
::查找圆形的方程,并给出以下信息。

center: $\begin{align*}(-3,-3)\end{align*}$ radius: 7
::半径: 7

center: $\begin{align*}(-7,6)\end{align*}$ radius: $\begin{align*}\sqrt{15}\end{align*}$
::中中-7,6)半径: 15

center: $\begin{align*}(8, -1)\end{align*}$ point on circle: $\begin{align*}(0, 14)\end{align*}$
::中心 : (8, - 1) 圆上的点 : (0, 14)

center: $\begin{align*}(-2, -5)\end{align*}$ point on circle: $\begin{align*}(3, 2)\end{align*}$
::中心-2,-5)圆上点3,2)

diameter endpoints: $\begin{align*}(-4, 1)\end{align*}$ and $\begin{align*}(6, 3)\end{align*}$
::直径端点-4,1)和(-6,3)

diameter endpoints: $\begin{align*}(5, -8)\end{align*}$ and $\begin{align*}(11, 2)\end{align*}$
::直径端点5,-8)和(11,2)

Is $\begin{align*}(-9, 12)\end{align*}$ on the circle $\begin{align*}(x+5)^2+(y-6)^2=54\end{align*}$ ? How do you know?
:- 9, 12) 是圆( x+5) 2+(y-6) 2=54? 你怎么知道 ?

Challenge Use the following steps to find the equation of the tangent line to the circle with center $\begin{align*}(3, -4)\end{align*}$ and the point of tangency $\begin{align*}(-1, 8)\end{align*}$ .
::挑战使用以下步骤寻找正切线与圆的方程(3,-4)和切点(-1,8)。

Find the slope of the radius from the center to $\begin{align*}(-1, 8)\end{align*}$ .
::查找半径的斜坡,从中间到(- 1, 8) 。

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。

Extension Rewrite the equation of the circle, $\begin{align*}x^2+y^2+4x-8y+11=0\end{align*}$ in standard form by completing the square for both the $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ terms. Then, find the center and radius.
::扩展名重写圆圆的方程式, x2+y2+4x-8y+11=0, 以标准格式填写, 以 x 和 y 条件填写正方形。然后, 找到中间和半径。

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

Circles Centered at (h,k) ::居于 (h,k) 的圆圆

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Circles Centered at (h,k)
::居于 (h,k) 的圆圆

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)