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
Eliminate from both equations.
Fernando Revilla
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 . 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.
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
You have a whole series of small errors- you are not being careful enough!
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 so y= 10- 3x, not 3x- 10.6x - 20 = 2y
y = 3x - 10
Of course once you have squared it the sign doesn't matter: (3x- 10)^2= (10- 3x)^2!Then taking equation (1) for substitution
x^2 + 2x + 1 + (3x - 10)^2 = 25
2x- 60x= -58x, not 62x.x^2 + 2x + 1 + 9x^2 - 60x + 100 = 25
10x^2 + 62x +101 = 25
76/2= 38, not 3610x^2 + 62x + 76 = 0
5x^2 + 31x + 36 = 010x^2- 58x+ 76= 0
etc
5x^2- 29x+ 38= 0
(x+ 2)(5x- 19)= 0.
Remember to use y= 10- 3x to find y, not y= 3x- 10.
The set of all points a distance of 5 units from (-1,0) is given by the equation
The set of all points a distance of 3 units from (2,1) is given by
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
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.