The line $-bx + ay + 2b = 0$ intersects circle on points A and B. Circle equation is $$(x-1)^2 + \big(y-\frac{a^2 + b^2 - a}{b}\big)^2 = \frac{(a^2 + b^2 - a)^2 + b^2}{b}$$ or after algebraic-transformations: $$bx^2 + by^2 - 2bx - 2(a^2 + b^2 - a)y =0 $$ Find the intersection points. One of them is $(2, 0)$. Thanks for any help for this hard task.
Trying mathematica :
$$x_1\to \frac{-2 b+\frac{2 a^2 b}{a^2+b^2}+\frac{a^3 b}{a^2+b^2}+\frac{a b^3}{a^2+b^2}-\frac{a \sqrt{32 b^2 \left(-a^2-b^2\right)+\left(4 a b+2 a^2 b+2 b^3\right)^2}}{2 \left(a^2+b^2\right)}}{b}$$
$$y_1\to \frac{4 a b+2 a^2 b+2 b^3-\sqrt{32 b^2 \left(-a^2-b^2\right)+\left(4 a b+2 a^2 b+2 b^3\right)^2}}{2 \left(a^2+b^2\right)}$$
$$x_2\to \frac{-2 b+\frac{2 a^2 b}{a^2+b^2}+\frac{a^3 b}{a^2+b^2}+\frac{a b^3}{a^2+b^2}+\frac{a \sqrt{32 b^2 \left(-a^2-b^2\right)+\left(4 a b+2 a^2 b+2 b^3\right)^2}}{2 \left(a^2+b^2\right)}}{b}$$
$$y_2\to \frac{4 a b+2 a^2 b+2 b^3+\sqrt{32 b^2 \left(-a^2-b^2\right)+\left(4 a b+2 a^2 b+2 b^3\right)^2}}{2 \left(a^2+b^2\right)}$$
But something is wrong because one solution must be (2,0).
EDIT:
This problem can be discribed by picture. In picture $O(1, \frac{a^2 + b^2 - a}{b})$, $A(2,0)$, $D(2a+2,2b)$ and $C(?,?)$. Equation of line and circle is in first line of my post.
Ah, an easier method just dawned on me.