# a question on properties of circle

• Nov 1st 2009, 12:30 AM
ukorov
a question on properties of circle
Referring to the attached pic, work out the length of OC - do not use trigonometric ratios stuff if possible.
• Nov 1st 2009, 12:51 AM
Opalg
Draw a tangent to the circle from C, say CT is the tangent and OT is the radius (so the lines CT and OT are perpendicular). There's a theorem which says that $\displaystyle CA\times CB = CT^2$, from which you can find CT. Then use Pythagoras.
• Nov 1st 2009, 01:09 AM
ukorov
Quote:

Originally Posted by Opalg
Draw a tangent to the circle from C, say CT is the tangent and OT is the radius (so the lines CT and OT are perpendicular). There's a theorem which says that $\displaystyle CA\times CB = CT^2$, from which you can find CT. Then use Pythagoras.

produce 2 similar triangles?
• Nov 1st 2009, 01:20 AM
ukorov
Quote:

Originally Posted by Opalg
Draw a tangent to the circle from C, say CT is the tangent and OT is the radius (so the lines CT and OT are perpendicular). There's a theorem which says that $\displaystyle CA\times CB = CT^2$, from which you can find CT. Then use Pythagoras.

but how do you know that TC is certainly a tangent of the circle???
actually if TC is produced from the original figure, angle ATC is does not appear to be a right angle.
• Nov 1st 2009, 01:33 AM
Opalg
Quote:

Originally Posted by ukorov
but how do you know that TC is certainly a tangent of the circle???
actually if TC is produced from the original figure, angle ATC is does not appear to be a right angle.

This is the picture that I intended. The angle OTC is a right angle.
• Nov 1st 2009, 01:49 AM
ukorov
Quote:

Originally Posted by Opalg
This is the picture that I intended. The angle OTC is a right angle.

oh i see but why is CA x CB = CT^2 ?
• Nov 1st 2009, 02:17 AM
aidan
Quote:

Originally Posted by ukorov
Referring to the attached pic, work out the length of OC - do not use trigonometric ratios stuff if possible.

Is it considered to be using trigonometric ratios
if you find the mid-point of the chord AB?

Since the chord length is given as 8, half would be 4.
If not then you have two 3-4-5 triangles inside the circle.
You then have sufficient information to complete the sides of a right triangle.
And from there, use pythagoras.

.
• Nov 1st 2009, 03:14 AM
ukorov
Quote:

Originally Posted by aidan
Is it considered to be using trigonometric ratios
if you find the mid-point of the chord AB?

Since the chord length is given as 8, half would be 4.
If not then you have two 3-4-5 triangles inside the circle.
You then have sufficient information to complete the sides of a right triangle.
And from there, use pythagoras.

.

well it is pretty easy to solve using tri ratios but this question was extracted from a textbook chapter on properties of circle.
• Nov 1st 2009, 05:50 AM
Opalg
Quote:

Originally Posted by ukorov
oh i see but why is CA x CB = CT^2 ?

Similar triangles. The triangles CTB and CAT are similar (angles are equal by the alternate segment theorem). Therefore $\displaystyle \frac{CA}{CT} = \frac{CT}{CB}.$