曲线 $(x-1)^2 + (y-3)^2 = 5$ 与直线 $y - 3x = 5$ 相交于两点。求这两个交点的坐标。

1. Express $y$ in terms of $x$ from the linear equation

From $y - 3x = 5$, we have:
$y = 3x + 5$

2. Substitute the expression for $y$ into the equation of the curve

Substitute $y = 3x + 5$ into $(x-1)^2 + (y-3)^2 = 5$:
$(x-1)^2 + (3x + 5 - 3)^2 = 5$
$(x-1)^2 + (3x + 2)^2 = 5$

3. Expand and simplify to form a quadratic equation in $x$

$(x^2 - 2x + 1) + (9x^2 + 12x + 4) = 5$
$10x^2 + 10x + 5 = 5$
$10x^2 + 10x = 0$

4. Solve the quadratic equation for $x$

$10x(x + 1) = 0$
$x = 0$ or $x = -1$

5. Find the corresponding $y$ values and state the coordinates

When $x = 0$, $y = 3(0) + 5 = 5$
When $x = -1$, $y = 3(-1) + 5 = 2$

The coordinates of the points of intersection are $(0, 5)$ and $(-1, 2)$.