# intersection of circles

• Mar 31st 2011, 12:44 PM
iragequit
intersection of circles
(x-3)^2 + (y-1)^2 = 49
(x+2)^2 + (y+4)^2 = 169

I expanded the two equations, etc etc, and I got 3 intersections, but the answer is 2 intersections (I even graphed the two circles and it's 2 intersections). I tried at least 3 times, looked over what I did 6 times, but I can't find out what I did wrong.. why am I getting 3 answers when there's 2 (2 of the 3 are right, the last one is wrong).
• Mar 31st 2011, 12:50 PM
SpringFan25
• Mar 31st 2011, 12:54 PM
iragequit
Quote:

Originally Posted by SpringFan25

(3,8) ; (10, 1) ; (-4, 1)

(-4, 1) is not an intersection.
• Mar 31st 2011, 12:56 PM
Quacky
You may be introducing an extraneous solution. This occurs frequently with squaring/rooting.

Let $x=4$
Then $x^2=16$.
But if you then square root both sides:
$x=\pm~4$

But $x\neq -4$ because it contradicts the original statement!

Do you see how an extra solution has arisen? The method in itself has no flaw, it's just that square rooting is a "one-to-many" operation. To verify solutions, you have to substitute into the original equations to check that they do indeed work. Please post your method and we can check for errors.
• Mar 31st 2011, 01:00 PM
iragequit
Quote:

Originally Posted by Quacky
You may be introducing an extraneous solution. This occurs frequently with squaring/rooting.

Let $x=4$
Then $x^2=16$.
But if you then square root both sides:
$x=\pm~4$

But $x\neq -4$ because it contradicts the original statement!

Do you see how an extra solution has arisen? The method in itself has no flaw, it's just that square rooting is a "one-to-many" operation. To verify solutions, you have to substitute into the original equations to check that they do indeed work. Please post your method and we can check for errors.

I substituted and I still got the right answer (and there were no square roots used). Here's my full solution.

(x-3)^2 + (y-1)^2 = 49 EQ. 1
(x+2)^2 + (y+4)^2 = 169 EQ. 2

Expanded EQ. 1 and 2 to get EQ. 3 and EQ. 4 respectively.

Subtracted EQ 4 from EQ 3 and got x = 11-y

Subsituted x = 11-y into EQ. 3 and got y = 8 and y = 1.

Substitued y = 8 into EQ. 3 and got x = 3.

Substitued y = 1 into EQ. 3 and got x = 10 and x = -4.

I hope that's not too hard to follow.
• Mar 31st 2011, 01:38 PM
SpringFan25
Quote:

Substitued y = 1 into EQ. 3 and got x = 10 and x = -4.
This is where the issue is. x=-4 is not a solution to your original system of equations. Its might be a solution, but it doesn't have to be.

You want the intersection of EQ 3. and the line y=11-x

Its true that an intersection occurs at y=1, but because EQ3 is a circle there are two possible x values at that y value. You need to check which one(s) you're interested in by seeing if they satisfy the original equations.

Graph on wolfram alpha
• Mar 31st 2011, 01:41 PM
iragequit
Quote:

Originally Posted by SpringFan25
This is where the issue is. x=-4 is not a solution to your original system of equations. Its might be a solution, but it doesn't have to be.

You want the intersection of EQ 3. and the line y=11-x

Its true that an intersection occurs at y=1, but because EQ3 is a circle there are two possible x values at that y value. You need to check which one(s) you're interested in by seeing if they satisfy the original equations.

Graph on wolfram alpha

Wait, what? I had x=11-y. How would I know when to change it to y=11-x?
• Mar 31st 2011, 01:45 PM
Quacky
Quote:

Originally Posted by iragequit
I substituted and I still got the right answer (and there were no square roots used). Here's my full solution.

(x-3)^2 + (y-1)^2 = 49 EQ. 1
(x+2)^2 + (y+4)^2 = 169 EQ. 2

Expanded EQ. 1 and 2 to get EQ. 3 and EQ. 4 respectively.

Subtracted EQ 4 from EQ 3 and got x = 11-y

Subsituted x = 11-y into EQ. 3 and got y = 8 and y = 1.

Substitued y = 8 into EQ. 3 and got x = 3.

Substitued y = 1 into EQ. 3 and got x = 10 and x = -4.

I hope that's not too hard to follow.

It would have been nicer if you'd replied with a full method so I wouldn't have had to have done every single calculation for myself.

3) $x^2+y^2-6x+-2y=39$
4) $x^2+4x+y^2+8y=149$

4)-3)
$10x+10y=110$
$x+y=11$
$y=11-x$

Sub. into:

1) $(x-3)^2 + (y-1)^2 = 49$
$
(x-3)^2 + (10-x)^2 = 49$

$x^2-6x+9+x^2-20x+100=49$

$2x^2-26x+60=0$

$x^2-13x+30=0$
$
(x-3)(x-10)=0$

$x=3$ or $x=10$
$
y=11-x$

$y=8$ or $y=1$

The extraneous solution is in this stage:
'Substitued y = 1 into EQ. 3 and got x = 10 and x = -4'

Because equation 3 is a polynomial involving squared terms, so there are possibly two solutions for any given x value that will satisfy equation 3, but that doesn't mean they'll satisfy equation 4. Notice how in my method I deliberately avoided this by using the $y=11-x$ instead of equation 3 for the last step? This was deliberately to avoid the extraneous solutions. Do you see why your method gave the extra solution?

$x^2+y^2-6x-2y=39$
If you let x=3,
$3^2+y^2-6(3)-2y=39$
$9+y^2-18-2y=39$
$y^2-2y-48=0$
$(y-8)(y+2)=0$
$y=8$ or $y=-2$
...are generated. Both solutions work for equation 3, but they don't necessarily work for equation 4. Why would they? You have only substituted into equation 3 after all!

So you can either check both in equation 4, or use the previously derived linear function to only obtain the valid solutions.
• Mar 31st 2011, 01:48 PM
Quacky
Quote:

Originally Posted by iragequit
Wait, what? I had x=11-y. How would I know when to change it to y=11-x?

You don't have to! Either will work.
• Mar 31st 2011, 01:57 PM
iragequit
Quote:

Originally Posted by Quacky
It would have been nicer if you'd replied with a full method so I wouldn't have had to have done every single calculation for myself.

3) $x^2+y^2-6x+-2y=39$
4) $x^2+4x+y^2+8y=149$

4)-3)
$10x+10y=110$
$x+y=11$
$y=11-x$

Sub. into:

1) $(x-3)^2 + (y-1)^2 = 49$
$
(x-3)^2 + (10-x)^2 = 49$

$x^2-6x+9+x^2-20x+100=49$

$2x^2-26x+60=0$

$x^2-13x+30=0$
$
(x-3)(x-10)=0$

$x=3$ or $x=10$
$
y=11-x$

$y=8$ or $y=1$

The extraneous solution is in this stage:
'Substitued y = 1 into EQ. 3 and got x = 10 and x = -4'

Because equation 3 is a polynomial involving squared terms, so there are possibly two solutions for any given x value that will satisfy equation 3, but that doesn't mean they'll satisfy equation 4. Notice how in my method I deliberately avoided this by using the $y=11-x$ instead of equation 3 for the last step? This was deliberately to avoid the extraneous solutions. Do you see why your method gave the extra solution?

$x^2+y^2-6x-2y=39$
If you let x=3,
$3^2+y^2-6(3)-2y=39$
$9+y^2-18-2y=39$
$y^2-2y-48=0$
$(y-8)(y+2)=0$
$y=8$ or $y=-2$
...are generated. Both solutions work for equation 3, but they don't necessarily work for equation 4. Why would they? You have only substituted into equation 3 after all!

So you can either check both in equation 4, or use the previously derived linear function to only obtain the valid solutions.

Wow.......... I could have just used the simple x=11-y........ WOW, how did I not think of that. So I lost a mark and wasted 5 minutes on a test doing absolutely non-sense. At least I learned, thanks guys.