Question

draw a circle with an equation of $x^{2}-4x + y^{2}+4y = 1$.

Explanation:

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

For \(x^{2}-4x\), we add \((\frac{-4}{2})^{2}=4\).
For \(y^{2}+4y\), we add \((\frac{4}{2})^{2}=4\).
The equation \(x^{2}-4x + y^{2}+4y=1\) becomes \((x - 2)^{2}-4+(y + 2)^{2}-4=1\).

Step2: Simplify the equation

Simplify \((x - 2)^{2}-4+(y + 2)^{2}-4=1\) to \((x - 2)^{2}+(y + 2)^{2}=9\).
The standard form of a circle is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center and \(r\) is the radius.
Here, the center is \((2,-2)\) and radius \(r = 3\).

Answer:

The circle has center \((2,-2)\) and radius \(3\). Plot the center \((2,-2)\) on the coordinate plane. Then, from the center, move \(3\) units up, down, left, and right to get four points on the circle: \((2,-2 + 3)=(2,1)\), \((2,-2-3)=(2,-5)\), \((2 - 3,-2)=(-1,-2)\), \((2+3,-2)=(5,-2)\). Sketch the circle passing through these points.