Formula for analytical finding ellipse and circle intersection points if exist I need a formula

Choose a coordinate system where the x axis is parallel to the ellipse major semiaxis, y axis parallel to the ellipse minor semiaxis, and origin where they intersect ("center of ellipse"). If a is the semi-major axis and b the semi-minor axis, and
${\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1tag\left\{1a\right\}label\left\{BtV1a\right\}}$
Assuming we have a nondegenerate ellipse (an ellipse with nonzero area), then ${\displaystyle a,b>0}$, and we can write above as
${\displaystyle x^2+\frac{a^2}{b^2}{y}^2-a^2=0tag\left\{1b\right\}label\left\{BtV1b\right\}}$
A circle of radius ${\displaystyle r\ge 0}$ centered at ${\displaystyle x=x_0,y=y_0}$ fulfills
${\displaystyle {(x-x_0)}^2+{(y-y_0)}^2=r^{2}tag\left\{2a\right\}label\left\{BtV2a\right\}}$
which we can also expand into ${\displaystyle x^2-2x_{0}x+y^2-2y_{0}y+x_0^2+y_0^2-r^2=0tag\left\{2b\right\}label\left\{BtV2b\right\}}$
To find the point (x,y) where the circle and the ellipse intersects, you need to solve the pair of equations, for example (1b) and (2b).
The system of equations has essentially form
$\{\begin{array}{rl}{x}^2+C_1{y}^2+C_2& =0\\ x^2+C_{3}x+y^2+C_{4}y+C_5& =0\end{array}\begin{array}{} \end{array}$
where ${\displaystyle C_1=\frac{a^2}{b^2},\sim C_2=-a^2,\sim C_3=-2x_0,\sim C_4=-2y_0,\sim C_5=x_0^2+y_0^2-r^2}$
One efficient way of solving (3) is to substract the first equation from the second, and solve for x. You'll find exactly one algebraic solution for x (that depends on y). Substitute it back into the first equation, and you have a quartic equation in y, which has zero, one, two, three, or four solutions. Solve that, then substitute the numeric value or values of y back into the algebraic solution for x, and you have the solution, which is either no intersection, one point of intersection, or two points of intersection.