Question

draw a circle with an equation of ( x^{2}+2x + y^{2}-2y = 7 ).

Explanation:

Step1: Complete the square for \(x\) terms

For \(x^{2}+2x\), we add \((\frac{2}{2})^{2}=1\).

$$x^{2}+2x=(x + 1)^{2}-1$$

Step2: Complete the square for \(y\) terms

For \(y^{2}-2y\), we add \((\frac{-2}{2})^{2}=1\).

$$y^{2}-2y=(y - 1)^{2}-1$$

Step3: Rewrite the circle equation

Substitute the completed - square expressions into the original equation \(x^{2}+2x+y^{2}-2y = 7\):

$$
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The standard form of a circle equation is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center and \(r\) is the radius. Here, \(a=-1\), \(b = 1\), and \(r = 3\).

Answer:

The center of the circle is \((-1,1)\) and the radius is \(3\). To draw the circle:

1. Locate the point \((-1,1)\) on the coordinate plane.
2. From the center \((-1,1)\), move \(3\) units up, down, left, and right. The points \((-1,1 + 3)=(-1,4)\), \((-1,1-3)=(-1,-2)\), \((-1-3,1)=(-4,1)\), and \((-1 + 3,1)=(2,1)\) are on the circle.
3. Sketch the circle passing through these four points.