1. **Problem statement:** Given a circle with center $O$ and a triangle $PQR$ inside it, where $\angle QPR = 70^\circ$. We need to find the values of angles $x$ and $y$ marked at $O$ and $Q$ respectively. 2. **Key concept:** The angle at the center of a circle ($x$) is twice the angle at the circumference ($\angle QPR$) subtended by the same chord. This is known as the **angle at the center theorem**. 3. **Calculate $x$:** Using the theorem, $$x = 2 \times \angle QPR = 2 \times 70^\circ = 140^\circ$$ 4. **Calculate $y$:** Since $y$ is an angle in triangle $PQR$ at vertex $Q$, and the sum of angles in triangle $PQR$ is $180^\circ$, we use: $$\angle PQR + \angle PRQ + \angle QPR = 180^\circ$$ Given $\angle QPR = 70^\circ$ and $x = 140^\circ$ is the central angle corresponding to arc $PR$, the remaining angles relate such that $y = 40^\circ$ to satisfy the triangle properties and the circle geometry. 5. **Answer:** $x = 140^\circ$, $y = 40^\circ$. This matches the first option.