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?

unnamed

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:

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

By Sheila Eastwood

Share this news article