Question

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

Explanation:

Step1: Complete the square for \(y\)

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.
Given \(x^{2}+y^{2}-2y = 8\).
For the \(y\) - terms: \(y^{2}-2y=(y - 1)^{2}-1\).
So the equation becomes \(x^{2}+(y - 1)^{2}-1=8\).

Step2: Rewrite in standard form

Add \(1\) to both sides of the equation \(x^{2}+(y - 1)^{2}-1=8\).
We get \(x^{2}+(y - 1)^{2}=9\).
Here, the center of the circle is \((0,1)\) (since \(a = 0\) and \(b = 1\)) and the radius \(r=\sqrt{9}=3\).

To draw the circle:

• Mark the center at the point \((0,1)\) on the coordinate plane.
• Using a compass (or by plotting points at a distance of \(3\) units from the center \((0,1)\) in all directions), draw the circle. For example, the points \((0,1 + 3)=(0,4)\), \((0,1-3)=(0,-2)\), \((3,1)\), \((-3,1)\) are on the circle.

Answer:

The circle has center \((0,1)\) and radius \(3\). Plot the center \((0,1)\) and then draw a circle with radius \(3\) units around it.