can you post your solution?
(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).
You may be introducing an extraneous solution. This occurs frequently with squaring/rooting.
Let
Then .
But if you then square root both sides:
But 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.
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.Substitued y = 1 into EQ. 3 and got x = 10 and x = -4.
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
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)
4)
4)-3)
Sub. into:
1)
or
or
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 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?
If you let x=3,
or
...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.