Chapter 9.2, Problem 30WE

To determine

To graph: The given inequality.

Answer to Problem 30WE

The center and the radius for the given inequality are $\left(-2,5\right)$ and $6$ .

Explanation of Solution

Given information: The given inequality is $x^2+y^2+4x-10y<7$ .

Formula used: ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation: Consider a circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ with center $\left(h,k\right)$ and radius $\left(r\right)$ .

Now, simplify the given inequality by making complete square,

  $\begin{array}{l}{x}^2+y^2+4x-10y<7\\ x^2+4x+4+y^2-10y+25<7+4+25\\ \left(x^2+4x+4\right)+\left(y^2-10y+25\right)<36\\ {\left(x+2\right)}^2+{\left(y-5\right)}^2<36\end{array}$

Since, $\left(-2,5\right):0^2+0^2<36$

Hence, the circle should be a solid line and the inside region should be shaded.

Compare ${\left(x+2\right)}^2+{\left(y-5\right)}^2<36$ to ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Hence, the center $\left(h,k\right)=\left(-2,5\right)$ and $r=6$ .

Therefore, the required graph of ${\left(x+2\right)}^2+{\left(y-5\right)}^2<36$ is given below.