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Chapter 1: Problem 1

Find the equation of the circle of radius 3 centered at: a) (0,0) b) (5,6) c) (-5,-6) d) (0,3) e) (0,-3) f) (3,0)

Short Answer

Expert verified
(a) \[x^2 + y^2 = 9\], (b) \[(x-5)^2 + (y-6)^2 = 9\], (c) \[(x+5)^2 + (y+6)^2 = 9\], (d) \[x^2 + (y-3)^2 = 9\], (e) \[x^2 + (y+3)^2 = 9\], (f) \[(x-3)^2 + y^2 = 9\].

Step by step solution

01

Understand the Standard Equation of a Circle

The standard equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. Our task is to find the equation of the circle using this formula with a given center and radius.
02

Applying the Center and Radius Values to the Equation

For each given center, \((h, k)\), use the formula \((x-h)^2 + (y-k)^2 = r^2\) with \(r = 3\). Substitute \(h\) and \(k\) with the given values for each part of the question.
03

Solve for Part a

The center is \((0, 0)\), so the equation is \[(x-0)^2 + (y-0)^2 = 3^2\] which simplifies to \[x^2 + y^2 = 9\].
04

Solve for Part b

The center is \((5, 6)\), so the equation is \[(x-5)^2 + (y-6)^2 = 3^2\] which simplifies to \[(x-5)^2 + (y-6)^2 = 9\].
05

Solve for Part c

The center is \((-5, -6)\), so the equation is \[(x+5)^2 + (y+6)^2 = 3^2\] which simplifies to \[(x+5)^2 + (y+6)^2 = 9\].
06

Solve for Part d

The center is \((0, 3)\), so the equation is \[(x-0)^2 + (y-3)^2 = 3^2\] which simplifies to \[x^2 + (y-3)^2 = 9\].
07

Solve for Part e

The center is \((0, -3)\), so the equation is \[(x-0)^2 + (y+3)^2 = 3^2\] which simplifies to \[x^2 + (y+3)^2 = 9\].
08

Solve for Part f

The center is \((3, 0)\), so the equation is \[(x-3)^2 + (y-0)^2 = 3^2\] which simplifies to \[(x-3)^2 + y^2 = 9\].

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Key Concepts

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

Standard Equation of a Circle
A circle in the coordinate plane can be represented by a specific algebraic expression. The **standard equation of a circle** is given by:\[(x-h)^2 + (y-k)^2 = r^2\]where:
  • (h, k) represents the center of the circle.
  • r represents the radius of the circle.
This equation captures the idea that any point \(x, y\) on the circle is equidistant from the center (h, k), with the specified distance being the radius r.
Let's illustrate this. Suppose our circle's center is at the origin, (0, 0), and the radius is 3. Plugging these values into the formula:\[(x-0)^2 + (y-0)^2 = 3^2\]This simplifies to:\[x^2 + y^2 = 9\]The resulting equation describes a circle centered at the origin with a radius of 3.
Center of Circle
The **center of a circle** is a crucial element in defining its position on the coordinate plane. It is represented by the coordinates (h, k) in the circle's equation:\[(x-h)^2 + (y-k)^2 = r^2\]
  • The x-coordinate of the center is h.
  • The y-coordinate of the center is k.
By simply adjusting h and k, we can change where the circle is located.
For example, if a center has coordinates (5, 6), the circle's equation becomes:\[(x-5)^2 + (y-6)^2 = r^2\]Here, replacing 3 for the radius:\[(x-5)^2 + (y-6)^2 = 9\]Even with different centers like (-5, -6) or (0, 3), we can swiftly integrate them into the equation without difficulty.
Radius of Circle
The **radius of a circle** is the distance from its center to any point on its circumference. In the standard circle equation:\[(x-h)^2 + (y-k)^2 = r^2\]r is the radius.
A circle's size changes according to this value, but not its shape. Increasing or decreasing the radius stretches or shrinks the circle, respectively.
Consider a circle with radius 3. In this instance, regardless of where the center is located, r remains 3. This means the equation will always look similar in form like so:\[(x-h)^2 + (y-k)^2 = 9\]where 9 is just 3 squared (32). This simplification makes it easy to understand and apply for various circle examples.

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