A Circle in a Cup

Let the circle be $x^2+(y-b)^2=5^2$ .

• Substitute $y=\frac{x^2}2$ :
• $x^2+(\frac{x^2}2-b)^2=5^2$
• $4x^2+(x^2-2b)^2=100$

• $x^4+(4-4b)x^2+(4b^2-100)=0$ has only one solution for $x^2$ .

• $\therefore \Delta=(4-4b)^2-4(4b^2-100)=0$

• $16(b-1)^2=16(b^2-25)$
• $b^2-2b+1=b^2-25$
• $2b-1=25$

• $b=13$

• $x^2=\frac{4b-4}2=24$

• $y=\frac{x^2}2=12$

• Therefore the required answer is $b-y$ which is $1$ .

The circle and parabola share two tangents. Since the centre of the circle is above the points of tangency, we use $y = -\sqrt{5^{2}-x^{2}}+c$ , where $c$ is a constant.

We equate the two derivatives:

$\frac{d}{dx} \frac{x^{2}}{2} = \frac{d}{dx} -\sqrt{5^{2}-x^{2}}+c$

$x = \frac{x}{\sqrt{25-x^{2}}}$ .

And solving for $x^{2}$ yields

$x^{2} = 24$ .

By Pythagoras' Theorem, $x^{2}+y^{2} = 5^{2}$ and hence $y = 1$ .