# question on properties of circle 3

• Nov 1st 2009, 08:30 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, 07:55 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 \$\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\$

• Nov 2nd 2009, 09:22 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, 10:42 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 \$\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 =\$ ?