Q. 66 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $65-70$, graph each equation and find the point(s) of intersection, if any.

The line $x+2y+6=0$ and the circle ${\left(x+1\right)}^2+{\left(y+1\right)}^2=5$. 

Step-by-Step Solution

Answer

The graph of the line $x+2y+6=0$ and the circle ${\left(x+1\right)}^2+{\left(y+1\right)}^2=5$ is,

From the graph, the two equations are intersecting at two points. They are, $\left(\frac{-16}{5},\frac{-7}{5}\right),\left(0,-3\right)$.

Step 1. Given information

$$\begin{gathered} x+2y+6=0 \\ {\left(x+1\right)}^2+{\left(y+1\right)}^2=5 \end{gathered}$$

First solve each equation for $y$.

Consider the first equation,

$$\begin{gathered} x+2y+6=0 \\ 2y=-x-6 \\ 2y=-\left(x+6\right) \\ y=-\frac{(x+6)}{2} \end{gathered}$$

Consider the second equation.

$$\begin{gathered} {\left(x+1\right)}^2+{\left(y+1\right)}^2=5 \\ {\left(y+1\right)}^2=5-{\left(x+1\right)}^2 \\ y+1=\pm \sqrt{5-{\left(x+1\right)}^2} \\ y=\pm \sqrt{5-{\left(x+1\right)}^2}-1 \end{gathered}$$

Step 2. Now sketch a graph for the obtained equations.

From the graph, the two equations are intersecting at the two points.

They are, $\left(\frac{-16}{5},\frac{-7}{5}\right),\left(0,-3\right)$.