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Chapter 13: Problem 16

For Problems \(15-32\), find the center and the length of a radius of each of the circles. $$ (x+6)^{2}+(y-9)^{2}=49 $$ \((-6,9) ; r=7\)

Short Answer

Expert verified
Center: (-6, 9); Radius: 7.

Step by step solution

01

Identify the General Form of a Circle's Equation

The general form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h,k)\) is the center of the circle and \(r\) is the radius.
02

Recognize the Given Circle's Equation

The given equation is \((x+6)^2 + (y-9)^2 = 49\). This follows the form \((x-h)^2 + (y-k)^2 = r^2\).
03

Identify Center Coordinates

By comparing the terms \((x+6)^2\) and \((y-9)^2\) to \((x-h)^2\) and \((y-k)^2\), we identify that \(h = -6\) and \(k = 9\). Therefore, the center of the circle is \((-6, 9)\).
04

Find the Radius

The equation \((x+6)^2 + (y-9)^2 = 49\) indicates that \(r^2 = 49\). Taking the square root of both sides, \(r = \sqrt{49} = 7\).
05

Conclusion

The center of the circle is \((-6, 9)\) and the radius is \(7\).

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Key Concepts

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

Center of Circle
To find the center of a circle given an equation in the standard form, we need to understand the general format of a circle's equation: \[(x-h)^2 + (y-k)^2 = r^2\]. Here, the center is represented by the coordinates \((h, k)\). When comparing this general equation to the specific circle equation \((x+6)^2 + (y-9)^2 = 49\):
  • The term \((x+6)^2\) shows that the horizontal component \(h\) is \(-6\) (notice the sign change).
  • The term \((y-9)^2\) indicates that the vertical component \(k\) is \(9\).
So, by interpreting these terms, we find that the center of the circle is \((-6, 9)\). This process involves reverse engineering the equation by identifying the values of \(h\) and \(k\), which are pivotal in pinpointing the center of any circle in such a form.
Radius of Circle
Understanding the radius of a circle from its equation involves recognizing the standard form \((x-h)^2 + (y-k)^2 = r^2\). In this equation, \(r^2\) represents the square of the radius. To find the actual radius:
  • Identify \(r^2\) from the given equation \((x+6)^2 + (y-9)^2 = 49\).
  • The equation shows that \(r^2 = 49\).
  • Take the square root of \(r^2\) to find \(r\): \(r = \sqrt{49} = 7\).
This approach derives the radius by reversing the squaring operation, helping us understand that the radius defines how far the circle stretches from its center, and for this particular equation, the radius is \(7\). Each unit on the grid or graph represents a linear distance, and the radius underlines how the circle broadens from its central point.
Standard Form of Circle
The standard form of a circle's equation is a compact way to express all the critical features of a circle, like its center and radius. This form is detailed as: \[(x-h)^2 + (y-k)^2 = r^2\]. Understanding this setup is essential when tackling circle-related problems.
  • The terms \((x-h)\) and \((y-k)\) help locate the center at \((h, k)\).
  • The \(r^2\) on the right side of the equation informs us of the radius. Here, it is essential to visualize \(r^2\) as the square of the radius, necessitating a final step to derive the radius by calculating its square root.
In the equation \((x+6)^2 + (y-9)^2 = 49\):- It matches the standard form as shown with \(h = -6\), \(k = 9\), and \(r^2 = 49\).This alignment helps clarify many geometrical properties of the circle, enabling us to understand and graph the circle effortlessly. Mastering this equation form opens the door to recognizing how alterations in \(h\), \(k\), or \(r\) impact the circle's position and size.

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Most popular questions from this chapter

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ \frac{(x-2)^{2}}{9}+\frac{(y-1)^{2}}{4}=1 $$

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For Problems \(1-30\), find the vertex, focus, and directrix of the given parabola and sketch its graph. $$ y^{2}-6 y-12 x+21=0 $$

For Problems \(15-32\), find the center and the length of a radius of each of the circles. Find the equation of the line that is tangent to the circle \(x^{2}+y^{2}+4 x-6 y-4=0\) \(x-4 y=3\) at the point \((-1,-1)\)

Find an equation of the ellipse that satisfies the oiven conditions. $$ \text { Foci }(\pm 1,0) \text {, length of minor axis is } 2 \quad x^{2}+2 y^{2}=2 $$
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