Chapter 10.9, Problem 33E

To determine

To find: find the coordinates of all vertices and determine whether the solution set bounds or not.

Answer to Problem 33E

  $\left(\sqrt{2},\sqrt{2}\right)$ and $\left(-\sqrt{2},-\sqrt{2}\right)$ are the vertices where the circle and the straight line intersect.

Explanation of Solution

Given:

  $\{\begin{array}{l}{x}^2+y^2\le 4\\ x-y>0\end{array}$

Calculation:

The given function

  $\begin{array}{l}{x}^2+y^2\le 4\\ x-y>0\end{array}$

Here. it is given that

  $\begin{array}{l}{x}^2+y^2\le 4\\ x-y>0\end{array}$

Then we take

  $x^2+y^2=\mathrm{4........}(1)$

Center of the circle $\left(0,0\right)$

Radius is $2$

Then the circle passes through points $(2,0),(-2,0),(0,2)$ , and $(0,-2)$

  $x-y>0$

We take $x-y=\mathrm{0........}(2)$

  $x=y$

This line passes through the origin

From equation $(2)$ , we have $y=x$

Substituting this expression into equation $(1)$

  $\begin{array}{l}{x}^2+x^2=4\\ 2x^2=4\\ x=\pm 2\\ x=\pm \sqrt{2}\end{array}$

When $x=\sqrt{2}$ , then $y=\sqrt{2}$

When $x=-\sqrt{2}$ , then $y=\sqrt{-2}$

  $\left(\sqrt{2},\sqrt{2}\right)$ and $\left(-\sqrt{2},-\sqrt{2}\right)$ are the vertices where the circle and the straight line intersect.

  

Conclusion:

Therefore, $\left(\sqrt{2},\sqrt{2}\right)$ and $\left(-\sqrt{2},-\sqrt{2}\right)$ are the vertices where the circle and the straight line intersect.