# Can your students solve this tricky puzzle?

Thursday 9th January 2020

Can you find the equation of a circle which touches the parabola $$y=x^2 at (0, 0)$$, but does not cross the parabola?

From symmetry, the centre of the circle must lie on the $$y$$ axis.

Let the centre of the circle be $$(a,0)$$

Equation of circle is $$x^2+(y-a)^2=a^2$$

Equation of parabola is $$y=x^2$$

At the point of intersection, substituting for $$y$$:

$y+(y-a)^2=a^2$

Collecting like terms and simplifying:

$y^2+y(1-2a)=0$

Since the circle touches but does not cross the parabola, this has only one root, so $$b^2-4ac=0$$

$$(1-2a)^2-4\times1\times0=0$$

$$a=\frac{1}{2}$$

The equation of the circle is:

$x^2+(y-\frac{1}{2})^2=\frac{1}{4}$