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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).


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.


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What's stopping you from using any of the techniques you've learned for solving a system of two equations in two variables? – Hurkyl Sep 16 '12 at 7:20
I have just edited my question. There's my answer :) – John Smith Sep 16 '12 at 7:41
The equation for your circle is wrong then:… – Hurkyl Sep 16 '12 at 9:21
Incidentally, once you have the equations right, there's a cute trick: once you eliminate $y$ and get a quadratic equation in $x$, you can divide out by $x-2$ because you know that's a solution already! (This would be more obvious if you eliminated $x$ first, I suppose) – Hurkyl Sep 16 '12 at 9:23
Thanks. It's preety obvious but i forgot about it :) Thanks a lot. Second point is x = 2(a^3 - a^2 + ab^2 + b^2)/(a^2+b^2) && y = b(2 - (4a)/(a^2 + b^2) ). – John Smith Sep 16 '12 at 9:30
up vote 2 down vote accepted

Ah, an easier method just dawned on me.

  • Construct the line L through O perpendicular to DA.
  • Find the point E where L intersects DA
  • The vector AC is twice the vector AE.
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