Chapter 13.1, Problem 6PSA

To determine

To verify: That the three points shown lie on the circle whose equation is $x^2+y^2=16$ .

Answer to Problem 6PSA

The points lie on the circle whose equation is $x^2+y^2=16$ .

Explanation of Solution

Given:

The figure is as follows.

  

Calculation:

To verify these points, substitute these points in the given equation and if the these points satisfy the equation then these points will lie on the circle of equation $x^2+y^2=16$ .

Take first point $\left(0,-4\right)$ .

Substitute 0 for $x$ and $-4$ for $y$ in the equation $x^2+y^2=16$ .

  $\begin{array}{c}{\left(0\right)}^2+{\left(-4\right)}^2=16\\ 16=16\end{array}$

Take second point $\left(4,0\right)$ .

Substitute $4$ for $x$ and 0 for $y$ in the equation $x^2+y^2=16$ .

  $\begin{array}{c}{\left(4\right)}^2+{\left(0\right)}^2=16\\ 16=16\end{array}$

Take third point $\left(0,4\right)$ .

Substitute 0 for $x$ and $4$ for $y$ in the equation $x^2+y^2=16$ .

  $\begin{array}{c}{\left(0\right)}^2+{\left(4\right)}^2=16\\ 16=16\end{array}$

All the points satisfy the equation.

Therefore, the points lie on the circle whose equation is $x^2+y^2=16$ .