## Precalculus (6th Edition) Blitzer

We know that both inequalities represent the circles centered at $\left( 4,-3 \right)$ with radius $2\sqrt{6}$. Let us consider the first inequality and put a test point $\left( 0,0 \right)$ in this inequality to check whether it satisfies the inequality or not. When we plug in $(0,0)$, we get $25\le 24$, which is incorrect. Therefore, the test point $\left( 0,0 \right)$ does not satisfy the inequality and the shaded region will be on and outside the circle. Also, consider the second inequality and put a test point $\left( 0,0 \right)$ in this inequality to check whether it satisfies the inequality or not as given below: \begin{align} & {{\left( x-4 \right)}^{2}}\text{+}{{\left( y+3 \right)}^{2}}\ge 24 \\ & 16+9\ge 24 \\ & 25\ge 24 \end{align} Which is correct. Therefore, the test point $\left( 0,0 \right)$ satisfies the inequality and the shaded region will be inside the circle. Thus, the coordinates lying on the circle are common in both the inequalities which means that there are infinitely many solutions.