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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?

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  • $\begingroup$ Because the center is $(x_1,y_1)$. Anyway your equation is wrong: all terms must be squared. $\endgroup$ – Aretino Sep 12 '15 at 16:38
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    $\begingroup$ Note that the set $\{(x,y) \in \mathbb R^2|(x-x_1)^2+(y-y_1)^2=r^2\}$ is circle with center in $(x_1,y_1)$ and radius $r$. $\endgroup$ – Zoran Loncarevic Sep 12 '15 at 16:40
  • $\begingroup$ I just edited the squares. $\endgroup$ – MrYouMath Sep 12 '15 at 16:40
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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.

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$x$ 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).

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  • $\begingroup$ Thank you for your answer but, am I not supposed to get two solutions? If coordinates of centers are known to me, and radii also, I would only get two coordinates from these solution. But two circles have two intersections, how to get that? $\endgroup$ – Alucard97 Sep 12 '15 at 17:06
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Subtract, freshman, subtract. The difference of the equations of two circles is a straight line. Plug that in one of the circular equations.

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