Q. 68 - Precalculus Enhanced with Graphing Utilities

Question

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

The circle ${\left(x+2\right)}^2+{\left(y-1\right)}^2=4$ and the parabola $y^2-2y-x-5=0$.

Step-by-Step Solution

Answer

The graph of the circle ${\left(x+2\right)}^2+{\left(y-1\right)}^2=4$ and the parabola $y^2-2y-x-5=0$ is,

From the graph, the intersecting points are, $\left(-3,\sqrt{3}+1\right),\left(-2,3\right),\left(-3,1-\sqrt{3}\right),\left(-2,-1\right)$.

Step 1. Given information

$$\begin{gathered} {\left(x+2\right)}^2+{\left(y-1\right)}^2=4 \\ {y}^2-2y-x-5=0 \end{gathered}$$

First solve each equation for $y$.

Consider the first equation,

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

Consider the second equation,

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

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

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