(x-x0)^2+(y-y0)^2=r^2 equation

Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2} = r^{2}$$
to
$$- r^{2} + \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) = 0$$
Expand the expression in the equation
$$- r^{2} + \left(\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2}\right) = 0$$
We get the quadratic equation
$$- r^{2} + x^{2} - 2 x x_{0} + x_{0}^{2} + y^{2} - 2 y y_{0} + y_{0}^{2} = 0$$
This equation is of the form
$$a\ x^2 + b\ x + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = - 2 x_{0}$$
$$c = - r^{2} + x_{0}^{2} + y^{2} - 2 y y_{0} + y_{0}^{2}$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(- 2 x_{0}\right)^{2} - 1 \cdot 4 \left(- r^{2} + x_{0}^{2} + y^{2} - 2 y y_{0} + y_{0}^{2}\right) = 4 r^{2} - 4 y^{2} + 8 y y_{0} - 4 y_{0}^{2}$$
The equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = x_{0} + \frac{\sqrt{4 r^{2} - 4 y^{2} + 8 y y_{0} - 4 y_{0}^{2}}}{2}$$
Simplify
$$x_{2} = x_{0} - \frac{\sqrt{4 r^{2} - 4 y^{2} + 8 y y_{0} - 4 y_{0}^{2}}}{2}$$
Simplify