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
Okay I will suggest another method.
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
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
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 =