SELECTION and TRAINING SESSION

We may assume that the parabola is defined by the equation $y=x^2$, while the circle is defined by the equation $(x-a{)}^2+(y-b{)}^2=R^2$. Let $A(a,b)$ be the center of the circle, $X_i(x_i,x_i^2)$, $i=1,2,3,4$, be common points of the circle and the parabola. Then $x_1,x_2,x_3,x_4$ are four roots of the equation $$(x-a{)}^2+(x^2-b{)}^2=R^2⟹$$ $$x^4-(2b-1)x^2-2ax+(a^2+b^2-R^2)=0.$$ It follows that $x_1+x_2+x_3+x_4=0$ (Vieta's formula). Denote by $M(k,l)$, $N(m,n)$ the coordinates of the midpoints of the arcs $X_1{X}_2$, $X_3{X}_4$, respectively. Then, in particular, $\overrightarrow{X_1{X}_2}\perp \overrightarrow{AM}$, which gives $$(x_1-x_2)(k-a)+(x_1^2-x_2^2)(l-b)=0\Rightarrow k-a=-(x_1+x_2)(l-b).$$ Since $M$ belongs to the circle, we have $(k-a{)}^2+(l-b{)}^2=R^2$ hence $$((x_1+x_2{)}^2+1)(l-b{)}^2=R^2,$$ so $(l-b{)}^2=R^2/((x_1+x_2{)}^2+1)$. Similarly, $$(n-b{)}^2=R^2/((x_3+x_4{)}^2+1).$$ Since $x_1+x_2=-(x_3+x_4)$, we get $$(l-b{)}^2=(n-b{)}^2.\text{\hspace{1em}(1)}$$ It is not difficult to see that the slope of the line $MA$ is positive (because that of the line $X_1{X}_2$ is negative) hence $b>l$; similarly, $b>n$. Therefore from (1) it follows that $b-l=b-n$, or $l=n$, which yields $MN\perp Oy$ as required.