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Chapter 12: Problem 29

$$\text {Sketch the graph of each equation.}$$ $$(x+1)^{2}+(y-1)^{2}=2$$

Short Answer

Expert verified
Circle with center \((-1, 1)\) and radius \( \root{2} \).

Step by step solution

01

- Identify the Equation Type

Recognize that the given equation \((x+1)^{2}+(y-1)^{2}=2\) is in the standard form of a circle equation \((x-h)^{2}+(y-k)^{2}=r^{2}\), where \(h\) and \(k\) are the coordinates of the center and \(r\) is the radius.
02

- Find the Center

Determine the center of the circle by comparing the given equation to the standard form. The center is at \((-1, 1)\) because \(h=-1\) and \(k=1\).
03

- Determine the Radius

Find the radius by taking the square root of the constant on the right-hand side of the equation. Since \(r^{2} = 2\), the radius \(r\) is \(\frac{\root{2}} \).
04

- Sketch the Circle

To sketch the circle, first plot the center at \((-1, 1)\). Then, draw a circle with radius \( \root{2} \) around this center. The circle will pass through points that are approximately 1.414 units away from the center in all directions.

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Key Concepts

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

Equation of a Circle
An equation of a circle in its standard form looks like this: \((x-h)^{2}+(y-k)^{2}=r^{2}\). Here, \(h\) and \(k\) represent the center coordinates of the circle, and \(r\) is the radius. To distinguish a circle's equation from other quadratic forms, look for the sum of squares of \x\ and \y\ terms, each shifted by some constants. If you think of the equation \((x+1)^{2}+(y-1)^{2}=2\), it matches the standard form. So, break it down: \(h=-1\), \(k=1\), and \(r^{2}=2\).
Center of a Circle
The center of a circle is quite straightforward to find once you understand the equation. In the standard form \((x-h)^{2}+(y-k)^{2}=r^{2}\), \(h\) and \(k\) are your center coordinates. For the equation \((x+1)^{2}+(y-1)^{2}=2\), compare and see \(x+1\) implies \h=-1\, and \(y-1\) means \k=1\. So, the center of this circle is at the point \(-1, 1\). This point is the middle of the circle, the exact place equidistant from every point on the circle's edge.
Radius of a Circle
The radius of a circle is the distance from its center to any point on the circle itself. In the equation \((x-h)^{2}+(y-k)^{2}=r^{2}\), \(r\) is the radius. To find it, simply take the square root of the right-hand side constant. In our case \((x+1)^{2}+(y-1)^{2}=2\), the constant is \2\. So, \(r\) is given by \sqrt{2}\. This means that every point on the circle is approximately \1.414\ units away from the center point \(-1, 1\). Visualize it by plotting this distance in all directions from the center.

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