1. **State the problem:** Show that the equation $4x^2 + 4y^2 + 7x - 11y = 0$ represents a circle. 2. **Recall the general form of a circle's equation:** $$ (x - h)^2 + (y - k)^2 = r^2 $$ where $(h,k)$ is the center and $r$ is the radius. 3. **Rewrite the given equation:** Divide the entire equation by 4 to simplify: $$ \cancel{4}x^2 + \cancel{4}y^2 + \frac{7}{4}x - \frac{11}{4}y = \cancel{0} $$ which simplifies to $$ x^2 + y^2 + \frac{7}{4}x - \frac{11}{4}y = 0 $$ 4. **Complete the square for $x$ and $y$ terms:** Group $x$ and $y$ terms: $$ (x^2 + \frac{7}{4}x) + (y^2 - \frac{11}{4}y) = 0 $$ For $x$: Take half of $\frac{7}{4}$, which is $\frac{7}{8}$, and square it: $$ \left(\frac{7}{8}\right)^2 = \frac{49}{64} $$ For $y$: Take half of $-\frac{11}{4}$, which is $-\frac{11}{8}$, and square it: $$ \left(-\frac{11}{8}\right)^2 = \frac{121}{64} $$ Add these squares to both sides: $$ (x^2 + \frac{7}{4}x + \frac{49}{64}) + (y^2 - \frac{11}{4}y + \frac{121}{64}) = 0 + \frac{49}{64} + \frac{121}{64} $$ 5. **Rewrite as perfect squares:** $$ \left(x + \frac{7}{8}\right)^2 + \left(y - \frac{11}{8}\right)^2 = \frac{170}{64} $$ Simplify the right side: $$ \frac{170}{64} = \frac{85}{32} $$ 6. **Conclusion:** The equation is now in the form $$ (x - h)^2 + (y - k)^2 = r^2 $$ with center $$ \left(-\frac{7}{8}, \frac{11}{8}\right) $$ and radius $$ r = \sqrt{\frac{85}{32}} $$ Therefore, the given equation represents a circle.