Circle theorems

**Geometor** · Dec 21st 2007, 08:43 AM

Hi, could anybody help me with this?
the question:
what is the distance from the centre (O) to C

if you need any clarification just ask
Any help appreciated!

**Plato** · Dec 21st 2007, 09:09 AM

Looking at the modified drawing, recall the secant theorem.
(AC)(BC)=(EC)(DC). Then solve for x.

**Soroban** · Dec 21st 2007, 09:28 AM

Hello, Geometor!

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

What is the distance from the centre (O) to C?

Let $x \:=\:OC.$

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

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

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

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

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

**Geometor** · Dec 21st 2007, 09:33 AM

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

Thanks for the help!

**JaneBennet** · Dec 21st 2007, 11:05 AM

Originally Posted by Soroban

Hello, Geometor!

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

Let $x \:=\:OC.$

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

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

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

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

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

That is also my method except that I didn’t use trigonometry. By Pythagoras’ theorem on triangle ODA, OD = $\sqrt{5^2-4^2}$ = 3 cm. By Pythagoras’ theorem on triangle ODC, OC = $\sqrt{(4+10)^2+3^2}=\sqrt{205}$ cm.

Originally Posted by Geometor

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

Thanks for the help!

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.

**Geometor** · Dec 21st 2007, 11:58 AM

haha thanks for pointing it out

i wrote 820 in my calculations though phew

**loui1410** · Dec 21st 2007, 03:04 PM

Originally Posted by Soroban

Hello, Geometor!

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

Let $x \:=\:OC.$

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

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

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

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

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

How do you know AD=4?

**JaneBennet** · Dec 21st 2007, 03:08 PM

Because OAB is an isosceles triangle (so the perpendicular from O to AB bisects AB).

**loui1410** · Dec 21st 2007, 03:09 PM

Oh right, sorry

it's 2:07 AM here lol

**Plato** · Dec 21st 2007, 04:25 PM

Originally Posted by JaneBennet

My recommendation: use the method that Soroban and I used.

Reading the title of the posting “Circle Theorems” why would you recommend against using a very basic theorem about circles? The approach that you advocate is not unique to theorems about circles but rather belongs to the general discussion about triangles.

**JaneBennet** · Dec 21st 2007, 05:04 PM

I didn’t realize you had to use the theorem to solve this problem. My apologies.

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