We can take WLOG parabola $y=ax^2$ for some $a\in R_+$. Let $P,Q,R,S$ be the points of intersection of circle centered at $O=(b,c)$ and the parabola such that $Q,R$ lie on the same side of the axis of the parabola and quadrilateral $PQRS$ is convex. First coordinate of points $P,Q,R,S$ will be denoted with small letter. For other two: $M=(m_x,m_y), N=(n_x,n_y)$ where $m_x\neq n_x$. We wish to prove $m_y=n_y$. Solving $y=ax^2\wedge (x-b)^2+(y-c)^2=r^2$ gives $a^2x^4+(1-2ac)x^2-2bx+b^2+c^2-r^2=0$ so Viete's formulas imply $p+q+r+s=0$. $$MO\perp SP\iff \vec{MO}\cdot\vec{SP}=0\iff m_x-b+a(p+s)(m_y-c)=0$$$$NO\perp QR\iff \vec{NO}\cdot\vec{QR}=0\iff n_x-b+a(q+r)(n_y-c)=0$$Adding and substracting equations yields $$m_x+n_x-2b=a(q+r)(m_y-n_y)\ \wedge\ m_x-n_x=a(q+r)(m_y+n_y-2c)$$$$(m_x-b)^2+(m_y-c)^2=(n_x-b)^2+(n_y-c)^2\iff (m_x-n_x)(m_x+n_x-2b)+(m_y-n_y)(m_y+n_y-2c)=0$$Thus $$(m_y-n_y)(m_y+n_y-2c)\left(1+a^2(q+r)^2\right)=0\iff (m_y-n_y)(m_y+n_y-2c)=0$$and $$(m_x-n_x)(m_x+n_x-2b)=0\iff m_x+n_x-2b=0$$Returning to $$m_x+n_x-2b=a(q+r)(m_y-n_y)$$we have $$m_y-n_y=0$$since $Q,R$ lie on the same side of the axis of the parabola so $q+r\neq 0$.