# Thread: Find the points where two circles cross

1. ## Find the points where two circles cross

I have two circles with equations:
(x - 1)^2 + (y - 4)^2 = 25 and (x - 9)^2 + (y - 4)^2 = 16

And I need to work out where the "lines" of the two circles cross. I can do this fine with a straight line and a circle but how do I do it with two circle equations?

I tried (y-4)^2 same so
(y - 4)^2 = 25 - (x - 1)^2
(y - 4)^2 = 16 - (x - 9)^2

and so 25 - (x - 1)^2 = 16 - (x - 9)^2

2. Expand this and solve the quadratic with x.

9 + (x^2 - 18x + 81) = (x^2 -2x + 1)

-18x + 90 = -2x +1
16x = 89

x = 89/16

Now that you've got the x value, find the y values.

3. But the circles cross in two places.

4. Originally Posted by angypangy
But the circles cross in two places.
Yes, but since the 2 circle centres lie on the line y=4,
the line y = 4 acts as an axis of symmetry for the graph.
Therefore the common chord going through the points of contact is vertical.
The value of x found will locate 2 y co-ordinates.

5. Hello, angypangy!

Find the intersections of: .$\displaystyle \begin{Bmatrix} (x - 1)^2 + (y - 4)^2 &=& 25 & [1] \\ (x - 9)^2 + (y - 4)^2 &=& 16 & [2]\end{Bmatrix}$

Subtract [1] - [2]: .$\displaystyle (x-1)^2 - (x-9)^2 \;=\;0$

. . . . $\displaystyle (x^2 - 2x+1) - (x^2 - 18x + 81) \;=\;9$

. . . . . . $\displaystyle x^2 - 2x + 1 - x^2 + 18x - 81 \;=\;9$

. . $\displaystyle 16x \:=\;89 \quad\Rightarrow\quad x \:=\:\frac{89}{16}$

Substitute into [1]: .$\displaystyle \left(\frac{89}{16}-1)^2 + (y-4)^2 \;=\;25$

. . . . . . . . . . . . . . . . . $\displaystyle (\frac{73}{16})^2 + (y-4)^2 \;=\;25$

. . . . . . . . . . . . . . . . . . . . . . .$\displaystyle (y-4)^2 \;=\;25-(\frac{73}{16})^2 \;=\; \frac{1071}{16^2}$

. . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle y - 4 \;=\;\pm\frac{\sqrt{1071}}{16}$

. . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle y \;=\;4 \pm\frac{\sqrt{1071}}{16}$

The intersections are: .$\displaystyle \left(\dfrac{89}{16},\;4 \pm \dfrac{\sqrt{1071}}{16} \right)$

6. Yes, just like a line can cross a circle at two points, in this problem, the circles cut each other at two points.

As Soroban showed you, the two coordinates are

$\displaystyle \left( \dfrac{89}{16}, 4+\dfrac{\sqrt{1071}}{16}\right)$

and

$\displaystyle \left( \dfrac{89}{16}, 4-\dfrac{\sqrt{1071}}{16}\right)$