Question

draw a circle with an equation of $x^{2}-2x + y^{2}=8$.

Explanation:

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

The general equation of a circle is \((x - a)^2+(y - b)^2=r^2\), where \((a,b)\) is the center and \(r\) is the radius.
For the given equation \(x^{2}-2x + y^{2}=8\), complete the square for the \(x\) - terms.
We know that \(x^{2}-2x=(x - 1)^{2}-1\) (using the formula \((m - n)^2=m^{2}-2mn + n^{2}\), here \(m = x\) and \(n = 1\)).
So the equation becomes \((x - 1)^{2}-1+y^{2}=8\).

Step2: Rewrite the equation in the standard form of a circle

Add \(1\) to both sides of the equation \((x - 1)^{2}-1+y^{2}=8\).
We get \((x - 1)^{2}+y^{2}=9\).
In the standard form \((x - a)^2+(y - b)^2=r^2\), here \(a = 1\), \(b = 0\) and \(r=\sqrt{9}=3\).

Answer:

The center of the circle is \((1,0)\) and the radius \(r = 3\). To draw the circle:

1. Plot the center point \((1,0)\) on the coordinate plane.
2. From the center, move \(3\) units up, down, left and right. The points \((1 + 3,0)=(4,0)\), \((1-3,0)=(-2,0)\), \((1,0 + 3)=(1,3)\) and \((1,0-3)=(1,-3)\) are on the circle.
3. Sketch a smooth curve passing through these four points and other points at an appropriate distance from the center \((1,0)\) with a radius of \(3\).