1. **Problem Statement:** Find the equation of the tangent line to the circle given by $$x^2 + y^2 - 3x + 10y = 15$$ at the point $(4, -11)$. 2. **Rewrite the circle equation in standard form:** The given equation is $$x^2 + y^2 - 3x + 10y - 15 = 0$$ which matches the general circle form $$x^2 + y^2 + 2gx + 2fy + c = 0$$ where $$2g = -3 \Rightarrow g = -\frac{3}{2}, \quad 2f = 10 \Rightarrow f = 5, \quad c = -15$$ 3. **Formula for tangent at point $(x_1, y_1)$ on the circle:** $$xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$$ 4. **Substitute values:** $$x_1 = 4, \quad y_1 = -11$$ So, $$4x - 11y + \left(-\frac{3}{2}\right)(x + 4) + 5(y - 11) - 15 = 0$$ 5. **Simplify step-by-step:** $$4x - 11y - \frac{3}{2}x - 6 + 5y - 55 - 15 = 0$$ Multiply all terms by 2 to clear denominator: $$2(4x) - 2(11y) - 3x - 12 + 10y - 110 - 30 = 0$$ which is $$8x - 22y - 3x - 12 + 10y - 110 - 30 = 0$$ 6. **Combine like terms:** $$ (8x - 3x) + (-22y + 10y) + (-12 - 110 - 30) = 0$$ $$5x - 12y - 152 = 0$$ 7. **Final answer:** The equation of the tangent line at point $(4, -11)$ is $$\boxed{5x - 12y - 152 = 0}$$