**Problem proposed by Peter Ash** (geometry forum 12 April 2006)

Let P be a point on a circle with center O, and let Q be a point on ray OP,
with OQ > OP. Let R and S be the points where the tangents from Q intersect
the circle, and let T and U be the midpoints of the segments RQ and SQ. Find
the locus traced out by T and U, as Q moves.

The curve looks something like y = x^{4}. Does it have a name? Can anyone
find a simple form of the equation for the curve?

*Solutions (both with Cabri -Geometry and calculating the equation)*

**Equation is obtained using the locus equation feature of Cabri-Geometry II
Plus.**

The figure is built like the **Luigi Tomasi**'s example in geometry forum:

the circle has equation: *x*^{2}+*y*^{2}=1, and the
point P has coordinates (1; 0).

Locus and equation:

**Calculating the equation** (by **François Rideau** and **Michel
Tixier**) with P (0;1)

Choose parameter t so that R coordinates are R(sin(t);cos(t)). So Q( 0;1/cos(t))
and T( (1/2) sin(t);(1/2)( cos(t) + 1/cos(t)) )

Then elimination of t via sin(t)^{2} + cos(t)^{2} = 1 gives
cartesian equation 4*x*^{4} + 4*x*^{2}.*y*^{2
}- 4*x*^{2 }-* y*^{2} + 1 = 0

**Alternate cartesian equation** (by **Michel Tixier**):

y = (1-2x^{2}) / √(1 - 4 x^{2}) with x in ] -1/2 , 1/2
[.

x = ±1/2 are two asymptotes of the curve.

Changing radius r = OP for 1,2,3,4,... we can "deduce" with Cabri
the general formula:

4*x*^{4 }+ 4*x*^{2}.*y*^{2 }- 4r^{2}.*x*^{2
}- r^{2}.* y*^{2 }+ r^{4} = 0

Calculating equation as above confirms this result.

Up to our knowledge this curve has not got a special name.

**Jean-Marie Laborde** proposes a generalization : *what happens if you choose
another point than the midpoint, e.g. a point at a constant ratio on QR*
?

**Joseph Hormière** found in the case r = 1 the equation x^{4} +
x^{2}.y^{2}– 2k.x^{2}– k^{2}.y^{2}
+ k^{2} = 0 where k is the ratio (k=1/2 gives Peter Ash's case).

**François Rideau** proposed another problem : *the construction of the
tangent to the quartic at point T* .