# Math Help - Finding the diameter of the circle

1. ## Finding the diameter of the circle

Write I have been given this problem which I simply cannot do? Any helpers please...

Here we go:

A, B and L are points on a circle, centre O
AB is a chord of the circle
M is the midpoint of AB
LOM is the midpoint of AB
AB = 24cm
LM = 18cm

What is the diameter of the circle?

2. Originally Posted by Natasha1
LOM is the midpoint of AB
LOM cannot be the midpoint of AB - the midpoint of a chord should be a point. Maybe you mean that LOM is the mediator (or perpendicular bisector) of AB?

3. Well spotted sorry I miss wrote it

A, B and L are points on a circle, centre O
AB is a chord of the circle
M is the midpoint of AB
LOM is a straight line
AB = 24cm
LM = 18cm

4. Originally Posted by Natasha1
Well spotted sorry I miss wrote it

A, B and L are points on a circle, centre O
AB is a chord of the circle
M is the midpoint of AB
LOM is a straight line
AB = 24cm
LM = 18cm
Draw a diagram, place the chord AB, the midpoint M and the centre of the circle O.
Since LOM is a straight line, LM is perpendicular to AB (being in fact the mediator)

Look at the triangle AOM (Right angle triangle):

OA = radius of the circle (R)

AM=24/2=12 cm

OM=LM-OL=18-R

Now apply Pythagoras Theorem in triangle AOM, this will lead you to an equation in R.

5. Hi natasha,
LM is the perpendicular bisector of AB
Let the distance from M to arc AB= x then 12 x12 = 18 x
Take it from there

6. This is what I get

r^2 = (18-r)^2 + 12^2
r^2 = 324 - 36r - r^2 + 144
r^2 = 468 -36r

What next?

7. Originally Posted by Natasha1
This is what I get

r^2 = (18-r)^2 + 12^2
r^2 = 324 - 36r - r^2 + 144
r^2 = 468 -36r

What next?
First line is ok, there is a mistake in the second one:

$r^2=324-36r+r^2+144$

$324-36r+144=0$

Can you solve it now and find r?

8. r = 13cm so diameter = 26cm. Where do you get all these braincells from?

9. Originally Posted by Natasha1
r = 13cm so diameter = 26cm. Where do you get all these braincells from?
Sometimes I wonder myself...lol.