Question

draw a circle with an equation of $x^2 + 4x + y^2 - 2y = 4$.

Explanation:

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

For \(x^{2}+4x\), we have \((x + 2)^{2}-4\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(2ab = 4x\Rightarrow b = 2\)).

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 \(a=y\), \(2ab=2y\Rightarrow b = 1\)).

Step3: Rewrite the circle equation

Substitute the completed - square forms into the original equation:
\((x + 2)^{2}-4+(y - 1)^{2}-1=4\)
\((x + 2)^{2}+(y - 1)^{2}=4 + 4+1\)
\((x + 2)^{2}+(y - 1)^{2}=9\)

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 \((-2,1)\) and the radius \(r = 3\).

To draw the circle:

1. Plot the center point \((-2,1)\) on the coordinate plane.
2. From the center, move 3 units up, down, left, and right. The points \((-2,1 + 3)=(-2,4)\), \((-2,1-3)=(-2,-2)\), \((-2-3,1)=(-5,1)\), \((-2 + 3,1)=(1,1)\) are on the circle.
3. Sketch the circle passing through these four points.

Answer:

Center \((-2,1)\) and radius \(r = 3\). Plot the center and use the radius to draw the circle on the given coordinate grid.