Recent content by drlisir

May I ask as c>0, so 1+4 r^2>0 Or 4 r^2 > -1 How can you deduce that r> 1/2?

16(c)I. Solving simultaneously: x^2 +(y-c)^2=r^2 and y=x^2 for the coordinates of the touching points, Hence : y+(y-c)^2=r^2, which is a quadratic equation in y. Buy the touching points are equal in y coordinates, so discriminant of the equation equals to 0. So (1-2c)^2 - 4 (c^2 - r^2 = 0...