Question 1132758
<pre>
First we find the points of intersection of the circle and
parabola by solving the system:

{{{system(x^2 + (y-c)^2=r^2,y=x^2)}}} 

We get this solution:

{{{x ="" +- sqrt("" +- expr(1/2)sqrt(4r^2 - 4c + 1) + c - 1/2)}}}

{{{y=x^2 = "" +- expr(1/2)sqrt(4r^2 -4c + 1) + c - 1/2)}}}

Since the circle touches the parabola in 2 places, like this:

{{{drawing(320,400,-2,2,-1,4,  
circle(0,1.3,.06),locate(0,1.3,"(0,c)"),
circle(0,1.3,1.02), graph(320,400,-2,2,-1,4,x^2) )}}} 

the value of c must be greater than it is in the case 
where all four solutions coincide and are equal at the origin (0,0). 
That is the case when y=0, and r=c

{{{drawing(320,400,-2,2,-1,4,  
circle(0,.5,.06),locate(0,.5,"(0,c)=(0,r)"),circle(0,0,.06),locate(0,0,"(0,0)"),
circle(0,.5,.5), graph(320,400,-2,2,-1,4,x^2) )}}} 

{{{y=x^2 = "" +- expr(1/2)sqrt(4r^2 -4c + 1) + c - 1/2)}}}

To find c in that case, we setting y=0 and r=c

{{{0= expr(1/2)sqrt(4c^2 -4c + 1) + c - 1/2)}}}

{{{0= expr(1/2)sqrt((2c-1)^2) + c - 1/2)}}}

{{{0= sqrt((2c-1)^2) + 2c - 1)}}}

{{{0= (2c-1) + 2c - 1)}}}

{{{0= 2c-1 + 2c - 1)}}}

{{{0= 4c - 2)}}}

{{{2 = 4c}}}

{{{2/4=c}}}

{{{1/2=c}}}

That is the case when c=r=1/2, therefore we must have c>1/2.

Edwin</pre>