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The line $x + y − 1 = 0$ intersects the circle $x^2 + y^2 = 13$ at $A(\alpha_1, \alpha_2)$ and $B(\beta_1, \beta_2)$. Without finding the coordinates of A and B, find the length of the chord AB.

Hint: Form a quadratic equation in $x$ and evaluate $|\alpha_1 − \beta_1 |$, and similarly find $|\alpha_2 − \beta_2 |$.

I formed the quadratic equation: $2x^2-2x-12=0$.

$|\alpha_1-\beta_1| = x$ length and $|\alpha_2-\beta_2| = y$ length right? Then calculate by distance formula.

I am not sure if am approaching it the right way since I get a weird equation. Any help please?

Race Bannon
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user255479
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3 Answers3

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Note that $$|\alpha_1-\beta_1|^2=(\alpha_1+\beta_1)^2-4\alpha_1\beta_1.$$ For $x^2-x-6=0$, by Vieta's formulas, you can have $$\alpha_1+\beta_1=-\frac{-1}{1}=1,\ \ \ \alpha_1\beta_1=\frac{-6}{1}=-6.$$

Another way is to use $$\overline{AB}=2\sqrt{r^2-d^2}$$ where $r$ is the radius of the circle and $d$ is the distance between the center of the circle and the line.

mathlove
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Hint: the difference of the roots of a quadratic equation $ax^2+bx+c=0$ is $$ |x_1-x_2|= \dfrac{\sqrt{\Delta}}{|a|} $$ as you can easely prove by the solution formula.

In your case you find:$\sqrt{\Delta}=5 \,,\,a=1 \Rightarrow |x_1-x_2|=|\alpha_1-\beta_1|=5$

Emilio Novati
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How about for a2 and b2? i tried subbing like a2 into euqation to get b2 but it doesn't correspond with the answer, which is 5(2^(1/2))

user254437
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