# question on properties of circle 3

• Nov 1st 2009, 09:30 AM
ukorov
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
• Nov 2nd 2009, 08:55 AM
Hello ukorov
Quote:

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 $QO_2$ to meet the circle at $R$. Then $\angle TBQ = \angle BRQ$ (alternate segment)

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

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

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

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

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

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

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

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

• Nov 2nd 2009, 10:22 AM
ukorov
Quote:

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?
• Nov 2nd 2009, 11:42 AM
Hello ukorov
Quote:

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

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