find the length of the chord 3x-y+9=0

of the circle x^2+y^2=5

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- Aug 19th 2009, 11:12 PM #1

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- Aug 19th 2009, 11:23 PM #2

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One can see a very dull method here.

Solve the two equations for the points of intersection,

$\displaystyle y=3x+9$

$\displaystyle x^2+y^2=5$

Therefore,

$\displaystyle x^2 + 9(x+3)^2=5$

$\displaystyle 10x^2 + 54x+81=5$

$\displaystyle 5x^2+27x+38=0$

so discriminant of this equation=729-760< 0

i.e. the line doesn't meet the given circle at all

I think you've made a typing mistake in your question.

- Aug 19th 2009, 11:23 PM #3
You have to solve the system

$\displaystyle \left\{\begin{array}{ll}x^2+y^2=5\\3x-y+9=0\end{array}\right.$

From the second ecuation $\displaystyle y=3x+9$. Replace y in the first equation. We get the quadratic

$\displaystyle 5x^2+27x+38=0$

But the discriminant is $\displaystyle \Delta =-31<0$, so the system has no real solution. That means the line doesn't intersect the circle.