Q .6 - Precalculus Enhanced with Graphing Utilities

Question

Graph the equation $x^2+y^2-4x+8y-5=0$.

Step-by-Step Solution

Answer

The required graph is  

Step 1. Given information.

Graph the equation $x^2+y^2-4x+8y-5=0$.

Step 2. Compare with equation of circle.

Equation of circle is:

${\left(x-p\right)}^2+{\left(y-q\right)}^2=r^2$

$\left(p,q\right)$ is the center of the circle, r is the radius.

Lets write $x^2+y^2-4x+8y-5=0$ in this form ${\left(x-p\right)}^2+{\left(y-q\right)}^2=r^2$.

$$\begin{gathered} x^2+y^2-4x+8y-5=0 \\ {x}^2-4x+y^2+8y-5=0 \\ {\left(x-2\right)}^2-4+{\left(y+4\right)}^2-16-5=0 \\ {\left(x-2\right)}^2+{\left(y+4\right)}^2=25 \end{gathered}$$

So, the center of circle has coordinates $\left(2,-4\right)$ and the radius is 5.

Step 3. Plot the graph of circle.

Plot the graph of circle ${\left(x-2\right)}^2+{\left(y+4\right)}^2=25$.