A circle is tangent to the parabola $y = \frac{x^2}{4}$ at $(2,1),$ and to the $y$-axis. Find the radius of this circle.
We can use calculus. For the parabola y = x^2/4, y' = 2x/4 = x/2. So for the point (a,a^2/4), the derivative is a/2. The slope of the normal is -2/a.
For the circle, (x - r)^2 + y^2 = r^2. Then by implicit differentation, 2(x - r) + 2y*2y/dx = 0. Setting dx = -a and dy = -2, and matching the parameters and solving, we get a = 2 + sqrt(3) and r = 7 - 3*sqrt(3).
So the radius of the circle is 7 - 3*sqrt(3).
Sorry your answer was incorrect, I have no idea about your solution but thanks for the effort
One question only; why is the vertical height from the x-axis to the circle centre \(1+\frac{r}{\sqrt{2}}\) ? Thanks!
+33616 Draw a horizontal line through point (2,1) until it is directly below the circle centre. Draw a vertical line from there to the circle centre. You now have a right angled, isosceles triangle with the radius as the hypotenuse (remembering that the hypotenuse here is at 45degrees ). Thus the two legs of this right angled triangle must be of length \(\frac{r}{\sqrt 2}\) by Pythagoras. Add this to the height (1) of the point (2,1).
