a question on properties of circle

ukorov · #1 $Old$ November 1st, 2009, 01:30 AM

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

Opalg · #2 $Old$ November 1st, 2009, 01:51 AM

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 $CA\times CB = CT^2$ , from which you can find CT. Then use Pythagoras.

ukorov · #3 $Old$ November 1st, 2009, 02:09 AM

Quote:

Originally Posted by Opalg $View Post$

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 $CA\times CB = CT^2$ , from which you can find CT. Then use Pythagoras.

produce 2 similar triangles?

ukorov · #4 $Old$ November 1st, 2009, 02:20 AM

Quote:

Originally Posted by Opalg $View Post$

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 $CA\times CB = CT^2$ , 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.

Opalg · #5 $Old$ November 1st, 2009, 02:33 AM

Quote:

Originally Posted by ukorov $View Post$

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.

ukorov · #6 $Old$ November 1st, 2009, 02:49 AM

Quote:

Originally Posted by Opalg $View Post$

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 ?

aidan · #7 $Old$ November 1st, 2009, 04:17 AM

Quote:

Originally Posted by ukorov $View Post$

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.

.

ukorov · #8 $Old$ November 1st, 2009, 05:14 AM

Quote:

Originally Posted by aidan $View Post$

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.

Opalg · #9 $Old$ November 1st, 2009, 07:50 AM

Quote:

Originally Posted by ukorov $View Post$

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 $\frac{CA}{CT} = \frac{CT}{CB}.$