# How to determine whether circles have points in common

Ok. So I have a question that gives me the equation of two circles stating that point $A$ and $B$ are the centre of these circles.

POINT $A$ is centre of circle: $(x-3)^2 + (y-11)^2 = 144$

POINT $B$ is centre of circle: $(x-12)^2 + (y+1)^2 = 9$

Part i) of the question asks to determine the length of $AB$, which I worked out using the distance formula to achieve $d= 15$ units.

Part ii) says "determine wether the circles have two points in common, just one point, or no points in common, and justify your answer."

This is where I'm lost and have no idea to work it out. I assume it may be something to do with $x$ and $y$ being equal in the two equations but then I don't know how to prove this.

Any help would be really appreciated. Thank you. Also, how do you know if $1$, $2$ or $0$ points intersect?

substracting from the equation (1) equation (2) we get $$x^2-6x+9+y^2-22y+121-x^2+24x-144-y^2-1-2y=135$$ this gives $$18x-24y=150$$ solve this equation for $x$ or $y$ and insert this in one equation of the given circles
$$x^2 - 6x + y^2 - 22y=14$$ and $$x^2 - 24x + y^2 + 2y=-136$$
subtracting these two then we get $$18x - 24y=150$$ $that implies$ $$x=\frac{75 + 12y}{9}$$ putting in one of the above then i am putting in first one then $$145y^2 + 330y - 3439=0$$ solve this or check only the descriminant Then you will know about the common points....