Q. 65 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

The graph of the line $x+2y=0$ 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, $\left(\frac{-6}{5},\frac{3}{5}\right),\left(2,-1\right)$.

Step 1. Given information

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

First solve each equation for $y$.

Consider the first equation,

$$\begin{gathered} x+2y=0 \\ 2y=-x \\ y=\frac{-x}{2} \end{gathered}$$

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.

In the graph, the two equations are intersecting at two points. They are, $\left(\frac{-6}{5},\frac{3}{5}\right),\left(2,-1\right)$.