Question

the circle ω touches the circle ω internally at p. the centre o of ω is outside ω. let xy be a diameter of ω which is also tangent to ω. assume p y > p x. let p y intersect ω at z. if y z = 2p z what is the magnitude of 6 p y x in degrees?

Answer 1

"Refer to the attached diagram in which we have drawn the common tangent TPR to the two given circles.

Let PZ=x so that YZ=2x. XY is contiguous to the small circle at Q. <ZQY=<ZPQ by the tangent chord or another part theorem.

But <ZQY = <PZQ - <ZYQ =<QPR - <RPX, once again by the aforesaid theorem.

Moreover, <QPR - <RPX = <QPX which means that PQ halves <YPX=90°; therefore <YPQ=45°. Now, YQ² =YZ*YP=2x*3x=6x² or YQ=√6*x.

Now, we use sine rule in triangle YPQ whereby, sin(YQP)/sin(45°) =3x/(√6*x) =√3/2. Apparently, <YQP is obtuse; hence <YQP=120 °. So, <PYX =180 -(120+45) =15°.

"

Answer 2

$\sf \angle pyx = 15 \degree$