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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).

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.

image1

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    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) ).2012-09-16

1 Answers 1

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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.