Chapter 1.5, Problem 101E

To determine

three points which lie on the graph of the equation $x^2+y^2=49$ .

Answer to Problem 101E

  $\left(0,7\right)$ , $\left(0,-7\right)$ And $\left(7,0\right)$

Explanation of Solution

Given:

The equation: $x^2+y^2=49$

To find three points which lie on the graph of the equation $x^2+y^2=49$ , substitute any three values for the variable x into the given equation and compute the corresponding y values.

First substitute $x=0$ in the given equation,

  $\begin{array}{c}{\left(0\right)}^2+y^2=49\\ 0+y^2=49\\ y^2=49\\ y=\pm 7\end{array}$

So, two points on the graph are $\left(0,7\right)$ and $\left(0,-7\right)$ .

Now substitute $x=7$ in the given equation,

  $\begin{array}{c}{\left(7\right)}^2+y^2=49\\ 49+y^2=49\\ y^2=49-49\end{array}$

  $\begin{array}{c}{y}^2=0\\ y=0\end{array}$

So, another point on the graph is $\left(7,0\right)$ .