1. **State the problem:** Find the solutions to the simultaneous equations: $$x^2 + y^2 = 16$$ $$y = 2x$$ These represent a circle centered at the origin with radius 4, and a line with slope 2. 2. **Substitute the linear equation into the circle equation:** Replace $y$ with $2x$ in the circle equation: $$x^2 + (2x)^2 = 16$$ 3. **Simplify the equation:** $$x^2 + 4x^2 = 16$$ $$5x^2 = 16$$ 4. **Solve for $x^2$:** $$x^2 = \frac{16}{5}$$ 5. **Find $x$ values:** $$x = \pm \sqrt{\frac{16}{5}} = \pm \frac{4}{\sqrt{5}} = \pm \frac{4\sqrt{5}}{5}$$ 6. **Find corresponding $y$ values using $y=2x$:** $$y = 2 \times \pm \frac{4\sqrt{5}}{5} = \pm \frac{8\sqrt{5}}{5}$$ 7. **Final solutions:** $$\left( \frac{4\sqrt{5}}{5}, \frac{8\sqrt{5}}{5} \right) \quad \text{and} \quad \left( -\frac{4\sqrt{5}}{5}, -\frac{8\sqrt{5}}{5} \right)$$ These are the points where the line intersects the circle.