# Thread: question on properties of circle 3

1. ## question on properties of circle 3

Referring to the attached pic:
(1) prove triangle BTQ ~ triangle QTA
(2) find TQ
(3) find the values of the radii

2. Hello ukorov
Originally Posted by ukorov
Referring to the attached pic:
(1) prove triangle BTQ ~ triangle QTA
(2) find TQ
(3) find the values of the radii
Produce $\displaystyle QO_2$ to meet the circle at $\displaystyle R$. Then $\displaystyle \angle TBQ = \angle BRQ$ (alternate segment)

$\displaystyle \Rightarrow \angle TBQ = \tfrac12 \angle BO_2Q$ (angle at centre)

$\displaystyle \angle TAO_1 = \angle TBO_2 = 90^o$ (angle between tangent and radius)

$\displaystyle \Rightarrow AO_1 \parallel BO_2$ (corresponding angles equal)

$\displaystyle \Rightarrow \angle AO_1T = \angle BO_2Q$ (corresponding angles)

But $\displaystyle \angle AQT =\tfrac12\angle AO_1T$ (angle at centre)

$\displaystyle \Rightarrow \angle AQT = \tfrac12\angle BO_2Q =\angle TBQ$ (proved)

So in $\displaystyle \triangle$'s $\displaystyle BTQ, QTA$:
$\displaystyle \angle AQT = \angle TBQ$ (proved)

$\displaystyle \angle ATQ = \angle QTB$ (same angle)
$\displaystyle \Rightarrow \triangle BTQ \sim \triangle QTA$

Hello ukorovProduce....
alright that makes sense now.
for question (2), is it to first prove that triangle AQB is right-angled, use Pythagarus Th. to find AB, and the rules of similar triangle to find TA, TO1, QO1, QO2?

4. Hello ukorov
Originally Posted by ukorov
alright that makes sense now.
for question (2), is it to first prove that triangle AQB is right-angled, use Pythagarus Th. to find AB, and the rules of similar triangle to find TA, TO1, QO1, QO2?
Yes. Note that $\displaystyle PA=PQ=PB$ (tangents from point to circle). So $\displaystyle Q$ lies on a circle that has $\displaystyle AB$ as a diameter. So $\displaystyle \angle AQB =$ ?

Hello ukorovYes. Note that $\displaystyle PA=PQ=PB$ (tangents from point to circle). So $\displaystyle Q$ lies on a circle that has $\displaystyle AB$ as a diameter. So $\displaystyle \angle AQB =$ ?