Question

draw a circle with an equation of $x^{2}+y^{2}-4y = 5$.

Explanation:

Step1: Complete the square for \(y\)

Given \(x^{2}+y^{2}-4y = 5\).
For the \(y\) - terms: \(y^{2}-4y=(y - 2)^{2}-4\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = y\), \(b = 2\)).
The equation becomes \(x^{2}+(y - 2)^{2}-4=5\).

Step2: Write in standard form

Add \(4\) to both sides of the equation \(x^{2}+(y - 2)^{2}-4 + 4=5 + 4\).
The standard form of the circle equation is \((x - 0)^{2}+(y - 2)^{2}=9\) (since the standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius).

The center of the circle is \((0,2)\) and the radius \(r=\sqrt{9}=3\).

To draw the circle:

• Locate the center at the point \((0,2)\) on the coordinate - grid.
• From the center, move \(3\) units up (to \((0,5)\)), down (to \((0,-1)\)), left (to \((-3,2)\)), and right (to \((3,2)\)).
• Sketch the circle passing through these four points.

Answer:

Center \((0,2)\) and radius \(r = 3\).