# Thread: intersection of circles

1. ## 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).

2. can you post your solution?

3. Originally Posted by SpringFan25
can you post your solution?
(3,8) ; (10, 1) ; (-4, 1)

(-4, 1) is not an intersection.

4. You may be introducing an extraneous solution. This occurs frequently with squaring/rooting.

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

But $\displaystyle x\neq -4$ 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.

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

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

But $\displaystyle x\neq -4$ 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.

6. 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

7. 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?

8. 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) $\displaystyle x^2+y^2-6x+-2y=39$
4) $\displaystyle x^2+4x+y^2+8y=149$

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

Sub. into:

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

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

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

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

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

$\displaystyle y=8$ or $\displaystyle y=1$

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 $\displaystyle y=11-x$ 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?

$\displaystyle x^2+y^2-6x-2y=39$
If you let x=3,
$\displaystyle 3^2+y^2-6(3)-2y=39$
$\displaystyle 9+y^2-18-2y=39$
$\displaystyle y^2-2y-48=0$
$\displaystyle (y-8)(y+2)=0$
$\displaystyle y=8$ or $\displaystyle y=-2$
...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.

9. 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.

10. 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) $\displaystyle x^2+y^2-6x+-2y=39$
4) $\displaystyle x^2+4x+y^2+8y=149$

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

Sub. into:

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

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

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

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

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

$\displaystyle y=8$ or $\displaystyle y=1$

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 $\displaystyle y=11-x$ 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?

$\displaystyle x^2+y^2-6x-2y=39$
If you let x=3,
$\displaystyle 3^2+y^2-6(3)-2y=39$
$\displaystyle 9+y^2-18-2y=39$
$\displaystyle y^2-2y-48=0$
$\displaystyle (y-8)(y+2)=0$
$\displaystyle y=8$ or $\displaystyle y=-2$
...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.