SOLUTION: The circle x^2 + (y-c)^2=r^2, where c >0 and r>0, lies inside the parabola y=x^2. The circle touches the parabola at exactly two points located symmetrically on opposite sides of t

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The circle x^2 + (y-c)^2=r^2, where c >0 and r>0, lies inside the parabola y=x^2. The circle touches the parabola at exactly two points located symmetrically on opposite sides of t      Log On


   

Question 1132758: The circle x^2 + (y-c)^2=r^2, where c >0 and r>0, lies inside the parabola y=x^2. The circle touches the parabola at exactly two points located symmetrically on opposite sides of the y-axis, as shown in the diagram. Deduce that c>1/2
so in the diagram, the parabola has vertex (0,0) and is positive for all values of x, and (o,c) is above the two points of intersection.
this was 2012 Q16c) HSC question
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
First we find the points of intersection of the circle and
parabola by solving the system:

system%28x%5E2+%2B+%28y-c%29%5E2=r%5E2%2Cy=x%5E2%29 

We get this solution:



y=x%5E2+=+%22%22+%2B-+expr%281%2F2%29sqrt%284r%5E2+-4c+%2B+1%29+%2B+c+-+1%2F2%29

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

 

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

 

y=x%5E2+=+%22%22+%2B-+expr%281%2F2%29sqrt%284r%5E2+-4c+%2B+1%29+%2B+c+-+1%2F2%29

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

0=+expr%281%2F2%29sqrt%284c%5E2+-4c+%2B+1%29+%2B+c+-+1%2F2%29

0=+expr%281%2F2%29sqrt%28%282c-1%29%5E2%29+%2B+c+-+1%2F2%29

0=+sqrt%28%282c-1%29%5E2%29+%2B+2c+-+1%29

0=+%282c-1%29+%2B+2c+-+1%29

0=+2c-1+%2B+2c+-+1%29

0=+4c+-+2%29

2+=+4c

2%2F4=c

1%2F2=c

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

Edwin