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 have \((x - 1)^{2}-1\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(b = 1\)).

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

For \(y^{2}+2y\), we have \((y + 1)^{2}-1\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\), \(b = 1\)).

Step3: Rewrite the circle equation

Substitute the completed - square forms into the original equation:
\((x - 1)^{2}-1+(y + 1)^{2}-1=7\).
Simplify to \((x - 1)^{2}+(y + 1)^{2}=9\).

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

Here, the center of the circle is \((1,-1)\) and the radius \(r = 3\).

To draw the circle:

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

Answer:

Center \((1,-1)\) and radius \(r = 3\). Plot the center and use the radius to draw the circle.