Analytic formula for intersection of two circles

Let be

$$(1)(x-a_0)^2+(y-b_0)^2=R^2_0=0;\:(x-a_1)^2+(y-b_1)^2=R^2_1$$

equations of two circles.We want to find the relative position of these two circle.
They could be disjoint, or tangent or have two common points.
All we have to do is to solve the system of equation.(1)
From the special form of our system written in the form

$$x^2+y^2-2a_0x-2b_0y+a_0^2+b_0^2-R_0^2=0$$
$$x^2+y^2-2a_1x-2b_1y+a_1^2+b_1^2-R_1^2=0$$

we have also

$$2(a_1-a_0)x+2(b_1-b_0)y+a_0^2+b_0^2-a_1^2-b_1^2+R_1^2-R_0^2=0$$

This is the equation of the radical axis, the line of points which have the same power with respect to the two circles.
Now we look to the system formed by one circle and the radical axis.

$$x^2+y^2-2a_0x-2b_0y+a_0^2+b_0^2-R_0^2=0$$
$$2(a_1-a_0)x+2(b_1-b_0)y+a_0^2+b_0^2-a_1^2-b_1^2+R_1^2-R_0^2=0$$

If radical axis is exterior to the circle the system has complex solutions and circles are disjoint.
If radical axis is tangent to the circle we have a double real solution and the corcles have a common tangent line,(they are tangent)
If the radical axis is secant to the circle the system has distinct real solutions and circle have two common points.

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