{(х-1)2+у>16 {х2+(у-2)2<9

To analyze the system of inequalities given: 1. \((x - 1)^2 + y > 16\) 2. \(x^2 + (y - 2)^2 < 9\) Let's interpret these inequalities geometrically. ### Inequality 1: \((x - 1)^2 + y > 16\) This inequality describes the region above the curve defined by \((x - 1)^2 + y = 16\). Rearranging gives: \[ y = 16 - (x - 1)^2 \] This represents a downward-opening parabola with its vertex at the point \((1, 16)\). The shaded region satisfying the inequality will be above this parabola. ### Inequality 2: \(x^2 + (y - 2)^2 < 9\) This inequality describes the interior of the circle defined by \(x^2 + (y - 2)^2 = 9\). The center of this circle is at the point \((0, 2)\) and the radius is \(3\) (since \(3^2 = 9\)). Therefore, the shaded region satisfying this inequality is the area inside the circle with the boundary excluded. ### Combined Inequalities To find the solution to the system, we look for the intersection of the two regions: 1. The region above the downward-opening parabola \(y = 16 - (x - 1)^2\). 2. The interior of the circle centered at \((0, 2)\) with radius \(3\). **Steps to solve graphically:** 1. Plot the parabola \(y = 16 - (x - 1)^2\). 2. Plot the circle with center \((0, 2)\) and radius \(3\), using the equation \(x^2 + (y - 2)^2 = 9\). 3. Identify the region where the area above the parabola overlaps with the area inside the circle. ### Solution Set - The solution set is the region that is both above the parabola \(y = 16 - (x - 1)^2\) and inside the circle defined by \(x^2 + (y - 2)^2 < 9\). - Finding the exact boundaries might require algebraic manipulation or plotting software for precise intersection points and region shading. If you need specific calculations or values for intersections, please let me know!