I am trying to compare the result of feasibility conditions for intersection of two custom circles, for when obtaining it geometrically and by using theorem of alternatives. My issue is that the dual problem gives another redundant constraint as well which is to avoid one circles being inside the other.
The two conditions for the circles are as follows (x,y are the variables): $$ (x - r_0)^2 + y^2 \leq r_0^2\\ (x - x_1)^2 + (y - y_1)^2 \leq r_1^2$$ So using geometry, the intersection is non-empty when the distance between center of circles is less than the sum of two radius. This gives the following feasibility condition: $$ (x_1 - r_0)^2 + y_1^2 \leq (r_1 + r_0)^2 $$
Then we try to solve the problem using the theorem of alternatives described in "Boyd and Vandenberghe, Convex Optimization" section 5.8.3 example of intersection of ellipsoids. The circles inequalities are written in the following standard quadratic format: $$ z^tz + 2[-r_0, 0].z \leq 0\\ z^t z + 2[-x_1, -y_1].z + (x_1^2 +y_1^2 - r_1^2)\leq 0 $$ Thus, we have the following parameters of dual function $$A(\lambda) = (\lambda_0 + \lambda_1)I $$ $$ b(\lambda) = \begin{bmatrix}-r_0 \lambda_0 - x_1 \lambda_1 \\ -y_1 \lambda_1 \end{bmatrix} $$ $$c(\lambda) = (x_1^2 + y_1^2 - r_1^2)\lambda_1$$ By the example of book, the dual problem is strong alternative of the original one. Therefore, having $$ \lambda \geq 0,\quad \lambda \neq 0, \quad -b^t(\lambda) A^{-1}(\lambda)b^t(\lambda) + c(\lambda) \geq 0$$ If the above problem is infeasible under certain conditions, then the original problem will be feasible under those conditions and vise versa. Substituting in the problem we have: $$ \lambda^t \begin{bmatrix} -r_0^2 & \frac{1}{2}(x_1^2 +y_1^2 - r_1^2 - 2 r_0x_1) \\ \frac{1}{2}(x_1^2 +y_1^2 - r_1^2 - 2 r_0x_1) & -r_1^2\end{bmatrix} \lambda \geq 0$$ If we want this to be infeasible, we must find conditions in which no $\lambda$ can satisfy this condition. It means that the negative of the matrix should be positive semi-definite. $$M:= \begin{bmatrix} r_0^2 & -\frac{1}{2}(x_1^2 +y_1^2 - r_1^2 - 2 r_0x_1) \\ -\frac{1}{2}(x_1^2 +y_1^2 - r_1^2 - 2 r_0x_1) & r_1^2\end{bmatrix} \succeq 0 $$. This eventually implies the following two conditions: $$ (x_1 - r_0)^2 + y_1^2 \leq (r_0 + r_1)^2$$ $$ (x_1 - r_0)^2 + y_1^2 \geq (r_0 - r_1)^2$$ The first condition is the same as what was obtained from the geometrical interpretation and is correct. The second condition however is a redundant check to not allow the circle to be inside the other. As Ivan mentioned, the circle being inside the other results in $M$ not being psd. This contradicts with the fact that the dual is strong alternative. Any suggestions are highly appreciated.