Chapter 7.1, Problem 52E

To determine

Tofind:The intersection point of the system of equations $\{\begin{array}{c}{x}^2+y^2=25\\ {\left(x-8\right)}^2+y^2=41\end{array}$ using graphical utility and verify the result by checking them in the original system.

Answer to Problem 52E

The intersection points of the system of equations $\{\begin{array}{c}{x}^2+y^2=25\\ {\left(x-8\right)}^2+y^2=41\end{array}$ are $\left(3,4\right)$ and $\left(3,-4\right)$ .

Explanation of Solution

Given information:

The given system of equations is $\{\begin{array}{c}{x}^2+y^2=25\\ {\left(x-8\right)}^2+y^2=41\end{array}$ .

Calculation:

Draw the graph of the equations $x^2+y^2=25$ and ${\left(x-8\right)}^2+y^2=41$ .

  

From the above graph, it can be seen that the intersection pointsare $\left(3,4\right)$ and $\left(3,-4\right)$ . So, the solution of the system of equations $\{\begin{array}{c}{x}^2+y^2=25\\ {\left(x-8\right)}^2+y^2=41\end{array}$ is $\left(3,4\right)$ and $\left(3,-4\right)$ .

Verify the solution by substituting the values of intersection point in the system of equations.

Substitute $3$ for $x$ and $4$ for $y$ in the first equation.

  $\begin{array}{c}{x}^2+y^2=25\\ {\left(3\right)}^2+{\left(4\right)}^2=25\\ 9+16=25\\ 25=25\end{array}$

Substitute $3$ for $x$ and $4$ for $y$ in the second equation.

  $\begin{array}{c}{\left(x-8\right)}^2+y^2=41\\ {\left(3-8\right)}^2+4^2=41\\ {\left(-5\right)}^2+16=41\\ 41=41\end{array}$

Now, it is clear that the intersection point satisfied both the equations.So, the ordered pair $\left(3,4\right)$ is a solution of the system of equations.

Substitute $3$ for $x$ and $-4$ for $y$ in the first equation.

  $\begin{array}{c}{x}^2+y^2=25\\ {\left(3\right)}^2+{\left(-4\right)}^2=25\\ 9+16=25\\ 25=25\end{array}$

Substitute $3$ for $x$ and $-4$ for $y$ in the second equation.

  $\begin{array}{c}{\left(x-8\right)}^2+y^2=41\\ {\left(3-8\right)}^2+{\left(-4\right)}^2=41\\ {\left(-5\right)}^2+16=41\\ 41=41\end{array}$

Now, it is clear that the intersection point satisfied both the equations. So, the ordered pair $\left(3,-4\right)$ is a solution of the system of equations.

Therefore, the intersection points of the system of equations $\{\begin{array}{c}{x}^2+y^2=25\\ {\left(x-8\right)}^2+y^2=41\end{array}$ are $\left(3,4\right)$ and $\left(3,-4\right)$ .