How to find intersection points of two circles

Question

I have two circles that have completely different center and radius, and I want to find their intersection points. However, everywhere on the internet in these cases it is assumed that they have the same center $x$ and $y$. All equations of circles seem to be in form $(x−x_1)^2+(y−y_1)^2=r^2$, and if there is another circle it is again in this form but $x$ and $y$ are not changed, only these $x_1$ and $y_1$ and $r$. How can that be?

Answer 1

The ($x_1 , y_1 $) here are the coordinates of the centre and r is the radius of the circle. The ($x,y$) are the coordinates of any point that satisfies the equation of the circle i.e. it lies on the circle. So if ($x_1,y_1$) are changed then centre is changed and if r is changed then radius is changed.

Answer 2

$x$ and $y$ are just variables.

The circle with center $(x_1,y_1)$ and radius $r$ is the set of points of the form $(x,y)$ where $x$ and $y$ are solutions to the equation $(x-x_1)^2+(y-y_1)^2=r^2$, where $r$ is the radius. This equation is determined by the values of $x_1$ and $y_1$ and $r$.

So if the first circle has center $(x_1,y_1)$ and radius $r$ we could let the second circle have center $(x_2,y_2)$ and radius $R$. Then the points that are on both circles are the points $(x,y)$ so that $x$ and $y$ simultaneously solve both of the following equations:

$(x-x_1)^2+(y-y_1)^2=r^2$ (this means $(x,y)$ is on the first circle)

$(x-x_2)^2+(y-y_2)^2=R^2$ (this means $(x,y)$ is on the second circle).

Answer 3

Subtract, freshman, subtract. The difference of the equations of two circles is a straight line. Plug that in one of the circular equations.