Hi, could anybody help me with this?

the question:

what is the distance from the centre (O) to C

http://img45.imageshack.us/img45/678/21870685kw9.png

if you need any clarification just ask

Any help appreciated!

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- Dec 21st 2007, 08:43 AMGeometorCircle theorems
Hi, could anybody help me with this?

the question:

what is the distance from the centre (O) to C

http://img45.imageshack.us/img45/678/21870685kw9.png

if you need any clarification just ask

Any help appreciated! - Dec 21st 2007, 09:09 AMPlato
Looking at the modified drawing, recall the secant theorem.

(AC)(BC)=(EC)(DC). Then solve for x. - Dec 21st 2007, 09:28 AMSoroban
Hello, Geometor!

I have a solution . . . hope it's acceptable.

Quote:

Draw an altitude from $\displaystyle O$ to $\displaystyle AB$; call it $\displaystyle OD.$

In right triangle $\displaystyle ODA,\:OA = 5,\:AD = 4\quad\Rightarrow\quad \cos A \:=\:\frac{4}{5}$

Law of Cosines: .$\displaystyle OC^2 \;=\;AC^2 + OA^2 - 2(AC)(OA)\cos A$

So we have: .$\displaystyle x^2 \;=\;18^2 + 5^2 - 2(18)(5)\left(\frac{4}{5}\right) \;=\;205$

Therefore: .$\displaystyle x\;=\;\sqrt{205}$

- Dec 21st 2007, 09:33 AMGeometor
thank you plato!

and soroban!

however i find using plato's secant theorem being the easier:

since:

(AC)(BC)=(EC)(DC).

(18)(10)=(10+x)(x)

=x^2+10x

x^2+10x-180=0

(-10 +/- sq.root 82) / 2

and since we need a positive value we do:

(-10 + sq.root 82) / 2

= 9.317821063....

and add the 5cm radius to get OC

= 14.3cm(1dp)

=sq. root 205 as soroban said :D

Thanks for the help! - Dec 21st 2007, 11:05 AMJaneBennet
That is also my method except that I didn’t use trigonometry. By Pythagoras’ theorem on triangle ODA, OD = $\displaystyle \sqrt{5^2-4^2}$ = 3 cm. By Pythagoras’ theorem on triangle ODC, OC = $\displaystyle \sqrt{(4+10)^2+3^2}=\sqrt{205}$ cm.

You made a mistake there; it should be 820, not 82. It’s also easier to make mistakes with your calculations using the secant method. My recommendation: use the method that Soroban and I used. :rolleyes: - Dec 21st 2007, 11:58 AMGeometor
haha thanks for pointing it out :D

i wrote 820 in my calculations though phew - Dec 21st 2007, 03:04 PMloui1410
- Dec 21st 2007, 03:08 PMJaneBennet
Because OAB is an isosceles triangle (so the perpendicular from O to AB bisects AB).

- Dec 21st 2007, 03:09 PMloui1410
Oh right, sorry :o it's 2:07 AM here lol

- Dec 21st 2007, 04:25 PMPlato
- Dec 21st 2007, 05:04 PMJaneBennet
I didn’t realize you had to use the theorem to solve this problem. My apologies.