1. **Problem statement:** We need to find the size of angle $x$ at the center $O$ of the circle. 2. **Given information:** - $O$ is the center of the circle. - Points $N, J, K, L, M$ lie on the circumference. - Angle at $N$ is $80^\circ$. - Angle at $K$ is $128^\circ$. - Angle $x$ is at the center $O$. 3. **Theorems and rules used:** - The sum of angles around point $O$ is $360^\circ$. - The central angle $x$ is equal to the sum of the arcs it intercepts. - The sum of interior angles of the polygon formed by points on the circle is $(n-2) \times 180^\circ$ where $n$ is the number of vertices. 4. **Step-by-step solution:** - Since $N, J, K, L, M$ lie on the circle, the polygon $NJKLM$ is cyclic. - The sum of interior angles of pentagon $NJKLM$ is $(5-2) \times 180^\circ = 540^\circ$. - We know two angles: $\angle N = 80^\circ$ and $\angle K = 128^\circ$. - Let the other three angles be $\alpha, \beta, \gamma$. - Then, $80 + 128 + \alpha + \beta + \gamma = 540$. - The central angle $x$ intercepts arcs corresponding to some of these angles. - Since the problem does not provide the other angles explicitly, but $x$ is the central angle corresponding to the arc between points $N$ and $K$ (assuming from the description), we use the property: $$\text{Central angle } x = 2 \times \text{Inscribed angle subtending the same arc}$$ - The inscribed angle subtending the same arc as $x$ is $\angle N = 80^\circ$. - Therefore, $$x = 2 \times 80^\circ = 160^\circ$$ 5. **Final answer:** $$\boxed{160^\circ}$$