Find the points where two circles cross

**angypangy** · October 18th 2010, 09:45 AM

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

But could not seem to get any sensible answer. Can anyone please help me out?

**Unknown008** · October 18th 2010, 10:04 AM

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.

**angypangy** · October 19th 2010, 04:10 AM

But the circles cross in two places.

**Archie Meade** · October 19th 2010, 08:07 AM

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.

**Soroban** · October 19th 2010, 08:23 AM

Hello, angypangy!

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

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

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

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

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

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

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

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

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

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

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

**Unknown008** · October 20th 2010, 05:28 AM

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

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

and

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

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