# How to find where two circles intersect

#### 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

#### FernandoRevilla

MHF Hall of Honor
HallsofIvy

#### HallsofIvy

MHF Helper
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 $$\displaystyle x^2+ y^2$$. 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.

FernandoRevilla

#### 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

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

#### bjhopper

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

#### HallsofIvy

MHF Helper
You have a whole series of small errors- you are not being careful enough!
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
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.

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

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

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

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

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

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

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

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.

#### bjhopper

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

bjh

#### HallsofIvy

MHF Helper
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.