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Get the intersections of the line $y=x+2$ with the circle $x^2+y^2=10$

What I did:

$y^2=10-x^2$

$y=\sqrt{10-x^2}$ or $y=-\sqrt{10-x^2}$

$ x+ 2 = y=\sqrt{10-x^2}$

If you continue, $x=-3$ or $x=1$ , so you get 2 points $(1,3)$, $(-3,-1)$

But then, and here is where the problems come:

$x+2=-\sqrt{10-x^2}$

I then, after a while get the point $(-3\dfrac{1}{2}, -1\dfrac{1}{2}$) but this doesn't seem to be correct. What have I done wrong at the end?

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Personally, I'd not take the square root, but substitute for $y$ from the linear equation into the quadratic, which gives you a quadratic in $x$ and two solutions. –  Mark Bennet Oct 14 '12 at 17:02
    
Yes, I know, that is true, but what is wrong about my way? –  JohnPhteven Oct 14 '12 at 17:02
    
It involves more manipulation and therefore more chance to make a mistake. –  Mark Bennet Oct 14 '12 at 17:04

2 Answers 2

up vote 2 down vote accepted

If you do everything correct, the solution of $x+2=-\sqrt{10-x^2}$ would be almost the same - $(1,-3)$, $(-3,1)$. It's refer to intersaction of $y = -(x+2)$ and original circle.

By the way, your answer is correct.

I have no idea how you'd get the point $(-3\dfrac{1}{2}, -1\dfrac{1}{2})$.

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$x+2=-\sqrt{10-x^2}$. $x^2+4x+4 = -10+x^2$. $4x = -14$. $x = 3,5$. –  JohnPhteven Oct 14 '12 at 16:56
1  
@ZafarS - the square of $-\sqrt{10-x^2}$ is $10-x^2$ –  Mark Bennet Oct 14 '12 at 17:03
    
ahh, that's what I wanted, thanks. –  JohnPhteven Oct 14 '12 at 17:18

Let the intersection be $(a,b)$, so it must satisfy both the given eqaution.

So, $a=b+2$ also $a^2+b^2=10$

Putting $b=a+2$ in the given circle $a^2+(a+2)^2=10$

$2a^2+4a+4=10\implies a=1$ or $-3$

If $a=1,b=a+2=3$

If $a=-3,b=-3+2=-1$

So, the intersections are $(-3,-1)$ and $(1,3)$

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I am aware it could be done that way, but what is wrong mathematically with my method? The correction model gives your method too, but I want to know why this is incorrect. –  JohnPhteven Oct 14 '12 at 16:54
    
$x+2=-\sqrt{10-x^2}$ actually refers to the intersection of $y=-(x+2)$ and $x^2+y^2=10$, the points of intersection being $(-3,1),(1,-3)$ –  lab bhattacharjee Oct 14 '12 at 16:57

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