# intersection points of two circles

I am trying to find the points at which two circles intersect. The circles I am working with are: $$(1) \qquad x^2+y^2 = \frac{9}{4}$$ $$(2) \qquad (x-2)^2+y^2=\frac9 4$$ I am following this answer: https://math.stackexchange.com/a/418932/136870.  When I subtract (1) from (2), I get $x=2$ . Substituting $x\leftarrow$ 2 in (1) gives me an imaginary number, $\sqrt{-\frac 7 4}$ for y, but I know these circles intersect. When I graphed these circles, I see that the intersection points are both on the line $x=1$. I also set the left side of (1) set to be equal to the left side of (2) since both are equal to $9/4$, but I got the same answer. Not sure what is going wrong here.

• The geometric viewpoint of Brian M. Scott is much better. However, if we expand $(x-2)^2+y^2$, we get $x^2-4x+4$, and when we subtract we get $4x-4=0$, giving $x=1$. – André Nicolas Mar 15 '15 at 7:12

## 2 Answers

Instead of following a recipe, look at the geometry: in this problem it’s exceptionally simple.

We can tell immediately from the equations that the centre $C_1$ of the first circle is the origin, $\langle 0,0\rangle$, and the centre $C_2$ of the second is $\langle 2,0\rangle$. Moreover, each circle has radius $\sqrt{\frac94}=\frac32$. Since the radii are the same, the points of intersection will lie on the perpendicular bisector of the line segment $\overline{C_1C_2}$, which is the line $x=1$. Thus, you need only find the two points on the line $x=1$ that are $\frac32$ units away from $C_1$ and $C_2$; that’s a straightforward application of the Pythagorean theorem.

Alternatively, subtract (1) from (2) or vice versa: $(x-2)^2 - x^2 = 0 \Rightarrow -4x+4 = 0$. Can you continue?