Which property of a circle or triangle show that $PT^2$ − $PU^2$ = $QT^2$ equation of concentric circle holds?

Question

Qn: If PT and PU are tangents from P to two concentric circles, with T on the smaller, and if the segment PT meets the larger circle at Q, then $PT^2$ − $PU^2$ = $QT^2$
as shown here

Attempt1: Using formula on this picture I get PU^2$ = $PQ*(2QT)$. Which properties of circle or triangle can help me to prove equation above?

Answer 1

HINT: the equation can be written as

$$PT^2 − QT^2 = PU^2$$

and then

$$(PT +QT)(PT-QT)=PU^2$$

Now apply the rule explained on your second figure.

Answer 2

You extend $PT$ so that it cuts the big circle in $Q$ and $Q'$. Let $O$ be center of our circles. You thus have $OT$ perpendicular to $QQ'$ so $$Q'T=QT$$ You then apply your relation (power of the point $P$): $$PU^2 = PQ\times PQ' = (PT-QT)(PT+QT)=PT^2 - QT^2$$