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Theory [Intersection points of two circles]

Consider that, we have the equations of two circles: $$x^2+y^2+2g_1x+2f_1y+c_1=0$$ $$x^2+y^2+2g_2x+2f_2y+c_2=0$$ First check the distance between the two centers \((-g_1,-f_1)\) and \((-g_2,-f_2)\) which is: $$d=\sqrt{(g_1-g_2)^2+(f_1-f_2)^2}$$ Again, sum of the radii of the two circles also needs to be calculated: $$r=r_1+r_2=\sqrt{g_1^2+f_1^2-c_1}+\sqrt{g_2^2+f_2^2-c_2}$$ Now, we look for specific conditions and take necessary steps according to the validation:

Now, there are two processes to proceed if the circles actually intersect at least once.
The first process is subtracting the two equations of the circle to find the equation of common chord between them:
Equation of common chord is: $$ (g_1-g_2)x+(f_1-f_2)y+(c_1-c_2)=0 $$ For the sake of clarity, consider \(p=g_1-g_2\rightarrow g_1=p+g_2\) , \(q=f_1-f_2=\rightarrow f_1=q+f_2\) and \(r=c_1-c_2\rightarrow c_1=r+c_2\). Rewriting the previous equation, we get \(px+qy+r=0\) as the equation of common chord of the two circles. This equation must intersect both of the two circles at the intersecting points \(x_1,y_1\) and \(x_2,y_2\), since this equation is the equation of common chord. Therefore, through substitution, we arrive at: $$ x=-\frac{r+qy}{p} $$$$ \left({\frac{r+qy}{p}}\right)^2+y^2-2\frac{g_1}{p}(r+qy)+2f_1y+c_1=0 $$ Solving this quadratic equation leads to two values of \(y\), that we may use to find two intersection points (or perhaps one).
The second process is much faster, and obviously, less tedious than the first.

Here, we find the constant \(a\) of the two circles, as depicted in the figure, according to the equation: $$ a=\frac{r_1^2-r_2^2+d^2}{2d} $$ and the segment h: $$ h=\sqrt{r_1^2-a^2} $$ The process we are following here, has everything based on the circle with radius \(r_1\) on a major scale, since constant h is a function of \(r_1\). The circle with radius \(r_1\) need not be the larger circle of the two. If you want, you can relabel the circles and choose working with the circle of radius \(r_2\).
Point \(P_4\) indicates one of the intersection points of the circles. Point \(P_3\) is simply the orthogonal drawn from point \(P_4\) on the line joining the centers of two circles. We need \(P_3\) to reach to \(P_4\).
To find the point \(P_3\) in vector form (i.e. in the form of \(a\hat{i}+b\hat{j}\) ), we use the following formula: $$ P_3=P_1+\frac{a}{d}(P_2-P_1) $$ We can extract each of the components of vector from this formula and go onward: $$ x_3=x_1+\frac{a}{d}(x_2-x_1) $$$$ y_3=y_1+\frac{a}{d}(y_2-y_1) $$ Now, we can say that the slope of the line joining the points \((x_3,y_3)\) and \((x_4,y_4)\) is equal to the ratio of the length of the sides \(\frac{h}{d}\), which is in fact \(tan\theta\). Again, the line joining the points \((x_1,y_1)\) and \((x_2,y_2)\) is perpendicular to the line joining the points \((x_3,y_3)\) and \((x_4,y_4)\). Therefore, the first line must have a slope that is \(-1\) times the second. Therefore, we can write: $$ tan\hspace{0.1cm}\theta=\frac{h}{d}=\frac{x_4-x_3}{y_4-y_3}=\frac{y_2-y_1}{x_1-x_2} $$ We can further rewrite it as: $$ \frac{h}{d}=\frac{x_4-x_3}{y_2-y_1}=\frac{y_4-y_3}{x_1-x_2} $$ Considering the first two expressions, we arrive at: $$x_4=x_3+\frac{h}{d}(y_2-y_1)$$ And similarly, $$y_4=y_3-\frac{h}{d}(x_2-x_1)$$ We can find \(P_5\) in the similar way, that in fact, has a slope of \(-\frac{h}{d}\) instead of \(\frac{h}{d}\) [as the value of \(y\) decreases as we increase the value of \(x\), the slope must be negative of the previous this time]. Therefore, two more equations come to our hand: $$x_5=x_3-\frac{h}{d}(y_2-y_1)$$ $$y_5=y_3+\frac{h}{d}(x_2-x_1)$$ Note that, \(P_5\) is the image of \(P_4\) with respect to \(P_3\).