Deductions from combined equation of two circles 
Given combined equation of two circles
$$x^4+y^4+2x^2y^2-4x^3-4xy^2+2x^2+2y^2+4x-3=0$$

In order to find the distance between the centres and the number of common tangents to the circles,
one way would be to factorize the equation as
$$(x^2+y^2-1)(x^2+y^2-4x+3)=0$$
However I was hoping whether there was another way to calculate distance between centres and number of tangents using the coeffecients  of the combined equation?
In cases where such combined equations are not easily factorisable, using coeffecients would be easier .
Thus for two circles having equation S1=0 and S2=0 , if we are given (S1)(S2)=0 , then is there a way we can find the above without having to factorize it to (S1)((S2)?
 A: Assuming the general equation of a circle as
$$x^2+y^2+2gx+2fy+c=0$$
For a system of two circles, the combined equation will be
$$(x^2+y^2+2g_1x+2f_1y+c_1)(x^2+y^2+2g_2x+2f_2y+c_2)=0$$
$$\Rightarrow x^4+2x^2y^2+y^4+2(g_1+g_2)x^3+2(f_1+f_2)x^2y+2(g_1+g_2)xy^2+2(f_1+f_2)y^3+(4g_1g_2+c_1+c_2)x^2+(4f_1f_2+c_1+c_2)y^2+4(g_1f_2+f_1g_2)xy+2(g_2c_1+g_1c_2)x+2(f_2c_1+f_1c_2)y+c_1c_2=0$$
Now, if the given equation is of the form $$x^4+2x^2y^2+y^4+Ax^3+Bx^2y+Axy^2+By^3+Cx^2+Dy^2+Exy+Fx+Gy+H=0$$
Comparing, we can get certain relations which we can use further.
$$A=2(g_1+g_2)$$
$$B=2(f_1+f_2)$$
$$C=4g_1g_2+c_1+c_2$$
$$D=4f_1f_2+c_1+c_2$$
$$E=4(g_1f_2+f_1g_2)$$
$$F=2(g_2c_1+g_1c_2)$$
$$G=2(f_2c_1+f_1c_2)$$
$$H=c_1c_2$$
Distance between the circles is given by
$$(g_2-g_1)^2+(f_2-f_1)^2$$
$$=(g_2+g_1)^2+(f_2+f_1)^2-4(g_1g_2+f_1f_2)$$
$$=\frac{A^2}{4}+\frac{B^2}{4}-(C-H+D-H)$$
$$=\boxed{\frac{A^2}{4}+\frac{B^2}{4}-(C+D-2H)}$$
However, I dont know if it is possible to find no of common tangents fron the coefficients in general.
