circle: b1^2+(b-r)^2=r^2
line: b2=-2/10rb1+2r
point b is on the intersection of the circle and the line
what are the x and y coordinates of point B? x=b1 and y=b2

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

circle: b1^2+(b-r)^2=r^2 line: b2=-2/10rb1+2r point b is on the intersection of the circle and the line what are the x and y coordinates of point B? x=b1 and y=b2

This problem involves finding the intersection of a circle and a line, where point B lies on both the curve and the line. Using the equation of the circle and the line, substitution is applied to solve for the x and y coordinates of point B. Key concepts include the equation of a circle, the equation of a line, and solving by substitution.

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