the origin O is the midpoint of a chord PQ of the circle , show the gradient of the chord PQ is 2

could someone please show me how you would figure this out, thanks

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- Oct 18th 2009, 05:20 AMmarkchord
the origin O is the midpoint of a chord PQ of the circle , show the gradient of the chord PQ is 2

could someone please show me how you would figure this out, thanks - Oct 18th 2009, 05:29 AMnolanfan
Equation of chord of a circle when its midpoint is given is where S is the equation the circle.

Given equation is

Equation of a chord when its midpoint

is

Here midpoint is 0rigin

Therefore

Therefore equation of chord is

whose slope is -6/-3 =2 - Oct 18th 2009, 05:32 AMmark
i don't think i understand that, is that called differentiation? because i haven't started doing that yet. is there a more simple way to do it?

- Oct 18th 2009, 05:56 AMnolanfan
Okay I will suggest another method.

Let

y= mx be chord to circle

centre is C (6,-3)

Mid pt is T(0,0)

Slope of CT = 0-(-3)/0-6 = -1/2

Since CT is perpendicular to the chord, slope of the chord = m= 2

the equation becomes y = 2x

therefore gradient = 2 - Oct 18th 2009, 06:16 AMmark
i think there's something wrong here, the book asked find the length of the chord PQ and the answer was but the length of the centre to the origin is which i think is what you were talking about. the centre (6, -3) is labelled as C by the way, so i'm not sure if it has anything to do with PQ

- Oct 18th 2009, 06:56 AMnolanfan
Well, I made a mistake by taking the midpoint (0,0) to be P randomly. I've changed the midpoint now to T. I didn't notice that the ends of the chord were P and Q. So let T (0,0) be the midpoint of PQ. Now the length of CT is . But CTP is a right angled triangle. therefore,

By pythogarus theorem

where TP = 1/2 length of chord.

Also CP = radius =

Length of chord = 2 x TP =