1. **State the problem:** We have a straight line with gradient (slope) 1 that cuts the y-axis at (0, 7). This line is tangent to the circle defined by the equation $$x^2 + (y - 3)^2 = 8$$ at point K. We need to find the coordinates of point K. 2. **Write the equation of the line:** Since the slope is 1 and it crosses the y-axis at 7, the line equation is: $$y = x + 7$$ 3. **Substitute the line equation into the circle equation:** Replace $$y$$ in the circle equation with $$x + 7$$: $$x^2 + ((x + 7) - 3)^2 = 8$$ Simplify inside the parentheses: $$x^2 + (x + 4)^2 = 8$$ 4. **Expand and simplify:** $$x^2 + (x^2 + 8x + 16) = 8$$ Combine like terms: $$2x^2 + 8x + 16 = 8$$ Subtract 8 from both sides: $$2x^2 + 8x + 8 = 0$$ 5. **Divide the entire equation by 2 to simplify:** $$\cancel{2}x^2 + \cancel{2} \times 4x + \cancel{2} \times 4 = 0$$ becomes $$x^2 + 4x + 4 = 0$$ 6. **Recognize the quadratic:** $$x^2 + 4x + 4 = (x + 2)^2 = 0$$ 7. **Solve for x:** $$x + 2 = 0 \Rightarrow x = -2$$ 8. **Find y using the line equation:** $$y = x + 7 = -2 + 7 = 5$$ 9. **Conclusion:** The coordinates of the tangent point K are: $$\boxed{(-2, 5)}$$