1. **State the problem:** We are given the equation of a circle: $$2x^2 + 2y^2 - 8x + 5y + 10 = 0$$ and need to find its radius. 2. **Rewrite the equation:** Divide the entire equation by 2 to simplify the coefficients of $x^2$ and $y^2$: $$x^2 + y^2 - 4x + \frac{5}{2}y + 5 = 0$$ 3. **Complete the square:** Group $x$ and $y$ terms: $$x^2 - 4x + y^2 + \frac{5}{2}y = -5$$ 4. For $x$ terms: $$x^2 - 4x = (x^2 - 4x + 4) - 4 = (x - 2)^2 - 4$$ 5. For $y$ terms: Take half of $\frac{5}{2}$ which is $\frac{5}{4}$, square it to get $\left(\frac{5}{4}\right)^2 = \frac{25}{16}$: $$y^2 + \frac{5}{2}y = \left(y^2 + \frac{5}{2}y + \frac{25}{16}\right) - \frac{25}{16} = \left(y + \frac{5}{4}\right)^2 - \frac{25}{16}$$ 6. Substitute back: $$(x - 2)^2 - 4 + \left(y + \frac{5}{4}\right)^2 - \frac{25}{16} = -5$$ 7. Combine constants on the right: $$-4 - \frac{25}{16} = -\frac{64}{16} - \frac{25}{16} = -\frac{89}{16}$$ So, $$(x - 2)^2 + \left(y + \frac{5}{4}\right)^2 = -5 + \frac{89}{16} = -\frac{80}{16} + \frac{89}{16} = \frac{9}{16}$$ 8. **Identify the radius:** The equation is now in standard form: $$(x - h)^2 + (y - k)^2 = r^2$$ where $r^2 = \frac{9}{16}$. 9. Therefore, the radius is: $$r = \sqrt{\frac{9}{16}} = \frac{3}{4}$$ **Final answer:** The radius of the circle is $\frac{3}{4}$.