Q. 5 - Precalculus Enhanced with Graphing Utilities

Question

Discuss the equation $x^2+y^2-2x+4y-4=0$. Graph it. 

Step-by-Step Solution

Answer

The graph of the equation is

Step 1. Given Information

We are given the equation $x^2+y^2-2x+4y-4=0$ and we need to graph the equation.

Step 2. Simplifying the equation

The standard form of the equation of a circle is${(x-h)}^2+{(y-k)}^2=r^2$.

Simplifying the given equation we get:

$$\begin{gathered} \Rightarrow x^2+y^2-2x+4y-4=0 \\ \Rightarrow x^2-2x+1+y^2+4y+4=1+4+4 \\ \Rightarrow {(x-1)}^2+{(y+2)}^2=9 \\ \Rightarrow {(x-1)}^2+{(y+2)}^2={\left(3\right)}^2 \end{gathered}$$

Comparing the equation with the standard form we get the center of the circle is at $(1,-2)$ and radius is $3$units.

Step 3. Plotting the graph

The graph of the equation is: