Finding the intersection points of two circles algebraically

Question

I need to find the intersection points of two circles with equations:

$(x+1)^2+(y-1)^2=1$ and $(x-1)^2+(y+1)^2=4$.

I understand how to find the points by plotting the circles but I am unsure of how to algebraically. If the second equation was equal to $1$, we would be able to set them equal to each other solve but that is not the case. I expanded each binomial into a trinomial but am unsure of where to go from there.

Thank you.

Have you tried subtracting three from the second one and then setting them equal? — user141592, Oct 18 '14 at 18:13
Try adding them together to find $x^2+y^2$, and then substitute back to find x in terms of y. Then you can get a quadratic in y. — user84413, Oct 18 '14 at 18:20

score 1 · Answer 1 · answered Oct 18 '14 at 18:30

$$\begin{align*}&x^2-2x+y^2+2y=2\\{}\\ &x^2+2x+y^2-2y=-1\end{align*}$$

Substract:

$$-4x+4y=3$$

and this means all the intersection points of the circles are on the above line, so if we take one of them, say $\;\left(x\;,\;\;x+\frac34\right)\;$ , we can substitute back in one of the circles:

$$(x-1)^2+\left(x+\frac74\right)^2=4\iff2x^2+\frac32x=4-1-\frac{49}{16}\implies$$

Solve the quadratic to find the points (it doesn't look like a nice answer)

Finding the intersection points of two circles algebraically

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