Question 1071489
A circle with radius {{{R>0}}} and center at the origin (0,0)
has the equation
{{{x^2+y^2=R^2}}} .
A parabola, opening upwards, and with its vertex on the y-axis
has the equation
{{{y=ax^2+b}}} , with some constants {{{b}}} , and {{{a>0}}} .
Such a parabola has the y-axis as its axis of symmetry.
You could also say the the y-axis is an axis of symmetry for
the circle with radius {{{R}}} and center at the origin (0,0).
As a consequence of this y-axis symmetry, for every solution of this system
(intersection point of the parabola and circle),
its reflection across the y-axis will also be a solution.
The only way to get 3 solution is for the odd solution to be
its own reflection across the y-axis image,
meaning that the odd intersection point is on the y-axis.
The circle crosses the y-axis at points (0,R) and (0,-R).
One of those points must be a point of the parabola,
specifically the point with {{{x=0}}} ,
parabola vertex {{{V(0,b)}}} .
So, it is either {{{V(0,-R)}}} , or {{{V(0,R)}}} ,
but a parabola opening upwards with vertex {{{V(0,R)}}}
would have {{{y>R}}} for all points other than the vertex,
and would touch the circle only at that one point,
meaning that the system would have only one solution.
So, {{{b=-R}}} , {{{V(0,-R)}}} is the vertex,
and {{{y=ax^2-R}}} is the equation of the parabola.
The point {{{P(sqrt(7),3)}}} is a solution, so it is
a point of the circle and a point of the parabola.
Substituting into {{{x^2+y^2=R^2}}} , we get
{{{(sqrt(7))^2+3^2=R^2}}}
{{{7+9=R^2}}}
{{{16=R^2}}}
The equation for the circle is {{{highlight(x^2+y^2=16)}}} ,
and {{{16=R^2}}} --> {{{R=sqrt(16)}}} --> {{{highlight(R=4)}}}
Substituting that value,
and the coordinates of {{{P(sqrt(7),3)}}} into {{{y=ax^2-R}}} , we get
{{{3=a(sqrt(7))^2-4}}}
{{{3=7a-4}}}
{{{3+4=7a}}}
{{{7=7a}}} --> {{{7/7=a}}} --> {{{highlight(a=1)}}} ,
and the equation for the parabola is
{{{highlight(y=x^2-4)}}} 
 
So, the system the problem asks us to "reverse-engineer" for is
{{{highlight(system(x^2+y^2=16,y=x^2-4))}}} .