1. **State the problem:** We are given the circle equation $$4x^2 + 4y^2 - 8x - 12y + 1 = 0$$ and a point $$P(3, 2.5)$$ outside the circle. We need to find the length of the tangent from point $$P$$ to the circle. 2. **Rewrite the circle equation in standard form:** Divide the entire equation by 4 to simplify: $$x^2 + y^2 - 2x - 3y + \frac{1}{4} = 0$$ 3. **Complete the square for both $$x$$ and $$y$$ terms:** Group $$x$$ and $$y$$ terms: $$x^2 - 2x + y^2 - 3y = -\frac{1}{4}$$ Complete the square: $$x^2 - 2x + 1 - 1 + y^2 - 3y + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 = -\frac{1}{4}$$ Rewrite: $$\left(x - 1\right)^2 - 1 + \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} = -\frac{1}{4}$$ 4. **Simplify the constants:** $$\left(x - 1\right)^2 + \left(y - \frac{3}{2}\right)^2 = -\frac{1}{4} + 1 + \frac{9}{4} = \frac{9}{4}$$ So the circle has center $$C(1, \frac{3}{2})$$ and radius $$r = \sqrt{\frac{9}{4}} = \frac{3}{2}$$. 5. **Find the distance from point $$P$$ to the center $$C$$:** $$d = \sqrt{(3 - 1)^2 + \left(2.5 - \frac{3}{2}\right)^2} = \sqrt{2^2 + \left(1\right)^2} = \sqrt{4 + 1} = \sqrt{5}$$ 6. **Use the tangent length formula:** The length $$L$$ of the tangent from point $$P$$ to the circle is given by: $$L = \sqrt{d^2 - r^2}$$ Calculate: $$L = \sqrt{5 - \left(\frac{3}{2}\right)^2} = \sqrt{5 - \frac{9}{4}} = \sqrt{\frac{20}{4} - \frac{9}{4}} = \sqrt{\frac{11}{4}} = \frac{\sqrt{11}}{2}$$ **Final answer:** The length of the tangent from point $$P(3, 2.5)$$ to the circle is $$\frac{\sqrt{11}}{2}$$.