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Chapter 11: Problem 31

(a) find the center and radius, then (b) graph each circle. $$ (x-4)^{2}+(y+2)^{2}=16 $$

Short Answer

Expert verified
Center: \((4, -2)\), Radius: 4, Graph by placing center at \((4, -2)\) with radius 4.

Step by step solution

01

Identify the Standard Form of a Circle

Start by identifying that the equation for a circle in its standard form is \((x-h)^{2}+(y-k)^{2}=r^{2}\).
02

Extract the Center Coordinates \((h, k)\)

Compare \((x-4)^{2}+(y+2)^{2}=16\) with the standard form. The center \((h, k)\) can be determined by the values that shift the circle horizontally and vertically: \( h = 4 \) and \( k = -2 \). Therefore, the center is \((4, -2)\).
03

Determine the Radius \(r\)

Identify the radius by comparing to the standard form. Here, \(r^{2} = 16\), so solve for \(r\) by taking the square root: \( r = \sqrt{16} = 4 \).
04

Plot the Center on the Graph

Mark the point \((4, -2)\) as the center of the circle on the coordinate plane.
05

Draw the Circle with Radius 4

Use a compass or measure 4 units in all directions (up, down, left, right) from the center point \((4, -2)\) to sketch the circle accurately.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

standard form of a circle
A circle's equation in its standard form provides us with all the basic details needed to understand its properties. The standard form of a circle's equation is \((x-h)^{2}+(y-k)^{2}=r^{2}\). Here, \(h\) and \(k\) represent the coordinates of the center of the circle, and \(r\) is the radius. This format helps us quickly identify the location and size of the circle. For example, consider the equation \((x-4)^{2}+(y+2)^{2}=16\). By comparing this to the standard form, we can immediately see the values of \(h\), \(k\), and \(r\). Understanding this form is crucial for breaking down any circle equation.
center of a circle
The center of a circle is the point from which all points on the circle are equidistant. To find the center, we use the values of \(h\) and \(k\) from the circle's standard equation. For instance, in \((x-4)^{2}+(y+2)^{2}=16\), we see that \(h=4\) and \(k=-2\). Thus, the center of the circle is at \((4, -2)\). This point is crucial because everything about the circle, including its radius and position on the graph, is determined from this center.
radius of a circle
The radius of a circle is the distance from the center to any point on the circle. To find the radius from the standard form equation, we solve for \(r\) in \((x-h)^{2}+(y-k)^{2}=r^{2}\). For our example equation \((x-4)^{2}+(y+2)^{2}=16\), we see that \(r^{2}=16\). By taking the square root, \(r=4\). This tells us that the circle is consistently 4 units away from its center at all points. The radius is essential for graphing and understanding the circle's overall size.
graphing circles
Graphing a circle involves plotting its center and using its radius to draw the curve. Here's a simple step-by-step process:
1. Identify the center coordinates \((h, k)\). For \( (x-4)^{2}+(y+2)^{2}=16 \), the center is \((4, -2)\).
2. Mark this center point on your graph.
3. From the center, measure the radius (which we've found to be 4 units) in all directions: up, down, left, and right.
4. Draw the circle by connecting these points with a smooth curve.
This will give you a perfect circle centered at \((4, -2) \) with a radius of 4.

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Most popular questions from this chapter

Solve the system of equations by using graphing. $$ \left\\{\begin{array}{l} y=-1 \\ (x-2)^{2}+(y-4)^{2}=25 \end{array}\right. $$

Graph. $$ \frac{(y+4)^{2}}{25}-\frac{(x+1)^{2}}{36}=1 $$

(a) write the equation in standard form and (b) graph. $$ 4 x^{2}+25 y^{2}+8 x+100 y+4=0 $$

Graph. $$ \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 $$

Graph. $$ 4 y^{2}-9 x^{2}=36 $$
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