Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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: wolframalpha.com/input/… –  Hurkyl Sep 16 '12 at 9:21
1  
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

1 Answer 1

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.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.