Explicit solution of an QCLP

The objective is linear, therefore the minimum is on the boundary and the inequality can be exchanged to equality. Let us first consider that the matrix $\mathbf{A}$ is positive definite. \begin{align} &\min_{\mathbf{x}}~ \mathbf{c}^{\text{T}}\mathbf{x}\\ &\text{s.t.}~~\mathbf{x}^{\text{T}}\mathbf{Ax}=1 \end{align} Then \begin{align} \mathcal{L}&=\mathbf{c}^{\text{T}}\mathbf{x}+\lambda(\mathbf{x}^{\text{T}}\mathbf{Ax}-1),\\ \frac{\partial \mathcal{L}}{\partial \mathbf{x}}&=\mathbf{c}+2\lambda\mathbf{Ax}=\mathbf{0} \implies \mathbf{x}=-\frac{1}{2\lambda}\mathbf{A}^{-1}\mathbf{c}\\ \frac{\partial \mathcal{L}}{\partial \lambda}&=\mathbf{x}^{\text{T}}\mathbf{Ax}-1=0 \end{align} By inserting the solution the second Eq. we get $2\lambda=\sqrt{\mathbf{c}^{\text{T}}\mathbf{A}^{-1}\mathbf{c}}$ and finally $\mathbf{x}=\frac{-\mathbf{A}^{-1}\mathbf{c}}{\sqrt{\mathbf{c}^{\text{T}}\mathbf{A}^{-1}\mathbf{c}}}$, where $\mathbf{A}^{-1}$ is also positive definite. The question now is how the problem changes if we use general matrix $\mathbf{A}$. One problem is that $\mathbf{c}^{\text{T}}\mathbf{A}^{-1}\mathbf{c}$ is not guaranteed to be nonnegative, therefore $\lambda$ cannot be determined in this way (if we use pseudoinverse). We can find an example, in which the minimum is $-\infty$, because some components of $\mathbf{x}$ can be out of control of $\mathbf{x}^{\text{T}}\mathbf{Ax}=1$, for example by $n=2$ and $\mathbf{A}=\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]$.