Thread: How to find where two circles intersect

1. How to find where two circles intersect

Hello

If I have these two circles:
(x + 1)^2 + y^2 = 25

and

(x - 2)^2 + (y-1)^2 = 9

How do I work out where the circles (or if) intersect?

It is difficult to set eg y= due to the different terms of x.

Angus

2. Eliminate $x^2+y^2$ from both equations.

Fernando Revilla

3. Fernando Revilla got in ahead of me but he is exactly right. Go ahead and multiply out the squares, then subtract one equation from another to eliminate $x^2+ y^2$. That will leave a single linear equation in x, and y. Solve for y in that equation, then substitute into either of the quadratic equations.

4. Aha.

like this:

(1) x^2 + 2x + 1 + y^2 = 25

(2) x^2 - 4x + 4 + y^2 - 2y +1 = 9

subtract (1) from (2)

6x - 2y - 4 = 16

6x - 20 = 2y

y = 3x - 10

Then taking equation (1) for substitution

x^2 + 2x + 1 + (3x - 10)^2 = 25

x^2 + 2x + 1 + 9x^2 - 60x + 100 = 25

10x^2 + 62x +101 = 25

10x^2 + 62x + 76 = 0

5x^2 + 31x + 36 = 0

etc

5. Hi Angus,

I subtracted equation 2 from 1 and obtained linear equation y = -3x +10. This is the equation of the common chord which is perpendicular to the line of centers. The line of centers has a positive slope.

bjh

6. You have a whole series of small errors- you are not being careful enough!
Originally Posted by angypangy
Aha.

like this:

(1) x^2 + 2x + 1 + y^2 = 25

(2) x^2 - 4x + 4 + y^2 - 2y +1 = 9

subtract (1) from (2)

6x - 2y - 4 = 16
x^2- x^2= 0, 2x-(-4x)= 6x, and y^2- y^2= 0 but 0- (-2y)= 2y, not -2y
6x+ 2y- 4= 16
6x - 20 = 2y

y = 3x - 10
6x- 20= -2y so y= 10- 3x, not 3x- 10.

Then taking equation (1) for substitution

x^2 + 2x + 1 + (3x - 10)^2 = 25
Of course once you have squared it the sign doesn't matter: (3x- 10)^2= (10- 3x)^2!

x^2 + 2x + 1 + 9x^2 - 60x + 100 = 25

10x^2 + 62x +101 = 25
2x- 60x= -58x, not 62x.

10x^2 + 62x + 76 = 0

5x^2 + 31x + 36 = 0
76/2= 38, not 36

etc
10x^2- 58x+ 76= 0
5x^2- 29x+ 38= 0
(x+ 2)(5x- 19)= 0.

Remember to use y= 10- 3x to find y, not y= 3x- 10.

7. Originally Posted by angypangy
Hello

If I have these two circles:
(x + 1)^2 + y^2 = 25

and

(x - 2)^2 + (y-1)^2 = 9

How do I work out where the circles (or if) intersect?

It is difficult to set eg y= due to the different terms of x.

Angus
The set of all points a distance of 5 units from (-1,0) is given by the equation

$[x-(-1)]^2+[y-0]^2=5^2\Rightarrow\ (x+1)^2+y^2=25\Rightarrow\ x^2+y^2+2x-24=0$

The set of all points a distance of 3 units from (2,1) is given by

$(x-2)^2+(y-1)^2=3^2\Rightarrow\ x^2-4x+4+y^2-2y+1=9\Rightarrow\ x^2+y^2-4x-2y-4=0$

To find the points (x,y) that are both 5 units from (-1,0) and 3 units from (2,1)
then the "x" values are equal in both equations.
The "y" values are also equal.

0-0=0

$\left[x^2+y^2+2x-24\right]-\left[x^2+y^2-4x-2y-4\right]=x^2-x^2+y^2-y^2+2x+4x+2y-24+4=0$

$6x+2y-20=0\Rightarrow\ 3x+y-10=0$

However, the resulting equation is the set of all points equidistant from the centres.
This is because we obtain the same result by subtracting the equations of
other circles with these centres, but whose radii squared differ by the exact same amount.

If you wish, you can then find the point of intersection of this line with either circle, by again allowing the "x" values to be equal and the "y" values to also be equal.

8. Dear Hallsofivy,
There is a typo in your final equation.It should read x-2 not x +2.

bjh

9. Originally Posted by bjhopper
Dear Hallsofivy,
There is a typo in your final equation.It should read x-2 not x +2.

bjh
Yes, you are right:
10x^2- 58x+ 76= 0
5x^2- 29x+ 38= 0
(x- 2)(5x- 19)= 0

Thanks.