Q. 69 - Precalculus Enhanced with Graphing Utilities

Question

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

$y=\frac{4}{x-3}$ and the circle $x^2-6x+y^2+1=0$.

Step-by-Step Solution

Answer

The graph of the equation $y=\frac{4}{x-3}$ and the circle $x^2-6x+y^2+1=0$ is,

From the graph, the intersecting points are, $\left(5,2\right),\left(1,-2\right)$.

Step 1. Given information

$$\begin{gathered} y=\frac{4}{x-3} \\ {x}^2-6x+y^2+1=0 \end{gathered}$$

First solve each equation for $y$.

Consider the first equation,

$y=\frac{4}{x-3}$

Consider the second equation,

$$\begin{gathered} x^2-6x+y^2+1=0 \\ {y}^2+1=-x^2+6x \\ {y}^2=-x^2+6x-1 \\ y=\pm \sqrt{-x^2+6x-1} \end{gathered}$$

Step 2 Now sketch a graph for the obtained equations.

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