Chapter 7, Problem 71RE

To determine

To sketch: the graph of the inequality.

Explanation of Solution

Given information:

Given inequality

  ${\left(x-1\right)}^2+{\left(y-3\right)}^2<16$

Calculation:

Let us consider the following inequality

The graph of ${\left(x-a\right)}^2+{\left(y-b\right)}^2<r^2$ represents the point inside the circle center $\left(a,b\right)$ and radius $r$ .

  ${\left(x-1\right)}^2+{\left(y-3\right)}^2<16$

First sketch the curve corresponding to equation ${\left(x-1\right)}^2+{\left(y-3\right)}^2<16$ . The circle divides the plane into two one inside it and one outside it.

Let us take the point $\left(0,0\right)$ from inside plane, clearly this point is satisfies the inequality and the region inside the circle represents the inequality excluding the points on the circle.