We are given the equations of a circle and a line, and point $B$ is at their intersection. Let's solve for the coordinates of point $B$ where:

1. The equation of the circle is: $b_1^2 + (b_2 - r)^2 = r^2$

2. The equation of the line is: $b_2 = -\frac{2}{10}r b_1 + 2r = -\frac{1}{5}r b_1 + 2r$

We need to find the $x$-coordinate $b_1$ and the $y$-coordinate $b_2$ of point $B$. To do this, we substitute the line equation into the circle equation and solve for $b_1$.

Step 1: Substitute the line equation into the circle equation

Substitute $b_2 = -\frac{1}{5}r b_1 + 2r$ into the circle equation: $b_1^2 + \left(-\frac{1}{5}r b_1 + 2r - r\right)^2 = r^2$

Simplify the expression inside the square: $b_1^2 + \left(-\frac{1}{5}r b_1 + r\right)^2 = r^2$

Step 2: Expand the square term

$b_1^2 + \left(\left(-\frac{1}{5}r b_1\right) + r\right)^2 = b_1^2 + \left(-\frac{1}{5}r b_1 + r\right)^2$ [ b_1^2 + \left(-\frac{1}{5}r b_1 + r\right)^2 = b_1^2 [ (b)+ need clarification two