1. **State the problem:** Find the equation of the line passing through point A(1, 2) and the centre of the circle, given that the line $3y - x = 5$ is tangent to the circle at point A. 2. **Understand the problem:** The radius of the circle at the point of tangency is perpendicular to the tangent line. Therefore, the line from the centre of the circle to point A is perpendicular to the tangent line $3y - x = 5$. 3. **Find the slope of the tangent line:** Rewrite the tangent line in slope-intercept form: $$3y - x = 5 \implies 3y = x + 5 \implies y = \frac{1}{3}x + \frac{5}{3}$$ The slope of the tangent line is $m_t = \frac{1}{3}$. 4. **Find the slope of the radius (line from centre to A):** Since the radius is perpendicular to the tangent, its slope $m_r$ satisfies: $$m_r = -\frac{1}{m_t} = -3$$ 5. **Write the equation of the line passing through A(1, 2) with slope $-3$:** Using point-slope form: $$y - 2 = -3(x - 1)$$ Simplify: $$y - 2 = -3x + 3$$ $$y = -3x + 5$$ **Final answer:** The equation of the line passing through A and the centre of the circle is $$\boxed{y = -3x + 5}$$