Q. 67 - Precalculus Enhanced with Graphing Utilities

Question

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

The circle ${\left(x-1\right)}^2+{\left(y+2\right)}^2=4$ and the parabola $y^2+4y-x+1=0$.

Step-by-Step Solution

Answer

The graph of the parabola $y^2+4y-x+1=0$ and the circle ${\left(x-1\right)}^2+{\left(y+2\right)}^2=4$ is,

From the graph, the two equations are intersecting at two points They are, $\left(0,-\sqrt{3}-2\right),\left(1,0\right),\left(0,\sqrt{3}-2\right),\left(1,-4\right)$.

Step 1. Given information

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

First solve each equation for $y$.

Consider the first equation,

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

Next consider the second equation,

$$\begin{gathered} {\left(x-1\right)}^2+{\left(y+2\right)}^2=4 \\ {\left(y+2\right)}^2=4-{\left(x-1\right)}^2 \\ y+2=\pm \sqrt{4-{\left(x-1\right)}^2} \\ y=\pm \sqrt{4-{\left(x-1\right)}^2}-2 \end{gathered}$$

Step 2. Now sketch a graph by using the obtained equations.

From the graph, the two equations are intersecting at two points. They are, $\left(0,-\sqrt{3}-2\right),\left(1,0\right),\left(0,\sqrt{3}-2\right),\left(1,-4\right)$.